Chapter 04: Stoichiometry

Mole Definition: The mole is the amount of a substance which contains as many elementary entities as there are atoms in 0.012kg (12g) of carbon-12 Explanation • The elementary entities may be atoms molecules, ions, electrons and other particles. It is represented by n. • The number of moles of a substance can be calculated by dividing mass in grams by molar mass. The formula for number of moles is: m Given mass Number of moles = M Molar mass Examples: The examples of elementary entities equal to one mole are: 1 mole of water = 6.022 × 1023 molecules of H2O = 6:022 × 1023 atoms of Carbon 1 mole of 120 1 mole of NaCl = 6.022 × 1023 formula units of NaCt = 6.022 x 1023 ions of sodium 1 mole of Nat Avogadro's number Definition: The number of entities present in one mole of a substance is a constant number named Avogadro's Number, i.e. 6.022 × 1023 Did you Know? S.Q. What is Avogadro's number? Ans. Avogadro's number is a physical constant representing the molar number of entities. The exact value of it is 6.022141.79 × 1023 mol. In calculations we use the rounded off value 6.6.02 × 10^23 Explanation: It is represented by NA. This value is attributed to an Italian scientist Amedo Avogadro (1776-1856). Avogadro's number is a physical constant representing the molar number of entities. The exact value of 6.02214179 × 1023 mol. In calculations we use the rounded off value 6.6.02 × 10^23 Examples: = 1 mole of Na 23 g of sodium = 1 mole of CaO 56.1 g of CaO 98 g of H2SO4 = 1 mole of H2SO4 60 g of CO% = 1 mole of CO Molar Mass Definition: The mass of one mole of a substance (element, compound or ionic species) is equal to the atomic mass, molecular mass, formula mass or ionic mass of a substance when expressed in grams is known as molar mass. Unit: Its unit is g/mol. Explanation: It is represented by M. The molar mass is the sum of the masses of the component atoms. The molar masses of all the atoms, molecules, formula units and ions always fix. Examples: Molar mass of Mg = 24.0 g/mol Molar mass of C12H22011 =342 g/mol Molar mass of Ag2Cr04 = 332 g/mol = 95g/mol Molar mass of PO* Calculation of molar mass of CCl4: The mass of one mole of CCl4 can be found by adding the masses of carbon and chlorine present. Molar mass of CCl4 = Molar mass of one C + Molar mass of CI × 4 = 12.0 X 1+35.5 x 4 Molar mass of CCl4 = 12.0 + 142.0 = 154.0 g/mol Quick Check 4.1 (a) Calculate the molar mass of KMnO4. Ans. Molar mass of KMnO4 = 39 + 55 + 16 × 4 = 158g / mol (b) Calculate the number of moles in 0.23g of sodium. Ans. Mass of sodium = 0:23g Molar mass of sodium = 23g / mol No. of moles of sodium = n= ? m n= M 0.23 g = - 23g / mol n = 0.01mol (c) Calculate the mass of 1.5 moles of Ca (OH)2. Ans. No of moles of Ca (OH)2 = 1.5mol Molar mass of Ca (OH)2 = 40 + 1 × 2 + 16 × 2 = 40 + 2 + 32 = 74g / mol = 6.6.02 × 10^23 atoms = 6.6.02 × 10^23 formula units = 6.6.02 × 10^23 molecules = 6.6.02 x 10^23 ions Mass of Ca(OH)2=m = ? m n=. M m 1.5 = 74 m = 1.5 x 74 Mass of Ca(OH)2 = 111g (d) The given mass of KClO3 is 24.5 g. Calculate its number of moles. Ans. Mass of KClO3 = m = 24.5g Molar mass of KClO3 = 39 + 35.4 + 16 ×3 = 39+35.4+48 = 122.4g / mol No. of moles of KClO3 = n =? m n = M 24.5 ple 519 122.4 n = 0.200g./ mol (e) How many molecules are present in 1.75 g of H2O2? Ans. Mass of H202= 1.75g Molar mas of H202 = 34g / mol No. of Molecules = ? No. of Molecules = M x NA 1.75 g = 34g/ mol * 6.6.02 × 10^23 = 0.051 mol × 6.6.02 × 10^23 = 3.098 x 1022 Molecules No. of Molecules (f) How many atoms are present in 15 g of gold ring? Ans. Mass of gold = 15g Molar mass of gold = 197g / mol m - × 6.6.02 × 10^23 No. of atoms of gold = M 15g - × 6.02 x1023 No. of atoms of gold = - 197g / mol = 0.076 × 6.6.02 × 10^23 No. of atoms of gold = 4.58 × 1022 atoms. RELATIONSHIP BETWEEN MOLE, MOLAR MASS AND AVOGADRO'S NUMBER

Relationship between molar mass and Avogadro's number A sample of 12.0 grams of natural carbon contains the same number of atoms as 4.0 grams of natural helium. Both samples contain 1 mole of atoms i.e. 6.022 × 1023 Number of atoms: It is interesting to know that different masses of elements have the same number of atoms i.e. "Avogadro's number. 1.0 of hydrogen = 1 mol of hydrogen = 6.6.02 × 10^23 atoms of H. = 1 mol of Na 23.0g of sodium = 1 mol of U 238.0g of uranium Number of molecules: One mole of different compounds has different masses but the same number of molecules i.e. Avogadro's number. =6.6.02 × 10^23 molecules of water = 1 mol of water 18.0g of H2O 18.0g of glucose = 1 mol of glucose Number of formula units: The number of formula units in one mole of an ionic compound is always same i.e. Avogadro's number. = 1 mole of NaCl 58.5g of NaCl = 1 mole of CaCO3 = 6.6.02 × 10^23 formula units of CaCO3 100g of CaCO3 = 1 mole of AgzCrO4 = 6.6.02 x 10^23 formula units of AgzCrOa 332g of Ag2Cr04 Number of ions: The number of ions in one mole of different ionic species is always the same, i.e. Avogadro's number = 1 mole of SO4* 96.1 g of SO4 = 1 mole of NO3! 62.0g of NO3 Sodium is heavier than hydrogen An atom of sodium is 23 times heavier than an atom of hydrogen. In order to have equal number of atoms, sodium should be taken 23 times greater in mass than hydrogen. Relationship between moles and Avogadro's Number One can calculate the number of moles by dividing the number of particles by Avogadro's number. No. of particles of a substance Number of moles = Number of particles = Molar x mass Mass in grams = 6.6.02 x 10^23 atoms of Na = 6.6.02 × 10^23 atoms of U = 6.6.02 × 10^23 molecules of glucose = 6.6.02 × 10^23 formula units of NaCl = 6.6.02 × 10^23 ions of SO4 = 6.6.02 × 10^23 ions of NO3! Avogadro's Number Mass of substance x NA Molar Mass x Avogadro's number No. of Moles Particles Sample Problem 4.4 A sample of glucose contains 3.76 × 1024 molecules of glucose. What is the number of moles in this quantity? 3.76×104 molecules No. of moles of glucose = - 6.02 x102 molecules mol No. of moles of glucose = 6.25 moles Sample Problem 4.5 How many atoms are there in a sodium metal that contains 2.3g? Solution 2.3 - = 0.1 mol Number of moles of sodium =- 23.0 Number of atoms of sodium = Number of moles of sodium x NA = 0.1x6.02x1023 = 6.6.02 × 102 atoms Sample Problem 4.6 Juglone, is a dye and is produced from the husks of black walnuts. The formula for Juglone is a) Calculate the molar mass of Juglone. b) Calculate number of moles in 0.87g of a sample of juglone extracted from black walnut husks. ple 519 Solution free 11 (a) C10H603 Molar mass of CioH6O3 = 10xA, (C) +6×1.0A, (H) + 3x A, (O) = (10×12.0) + (6x1.0) + (3x16.0) Molar mass of C1oH,03 =120 + 6+48 = 174g/ mol Mass (b) Moles of juglone = Molar Mass 0.87 g = - = 0.005 moles 174 g/mol Interesting Information! S.Q. Give benefits of Juglone. Ans. Juglone, is a natural herbicide (weed killer). It kills off competitive plants around the black walnut tree but does not affect grass and other noncompetitive plants. Quick Check 4.2 (a) A copper wire contains 27.10 × 1025 atoms of copper. Calculate the number of moles of copper. Ans. No of atoms of Cu = 27.10 × 1025 Avogadro's Nümber = NA = 6.02 x1023 No of moles = n= ? 27.10x1.025 n= 6.02x1023 n = 4.50 moles Calculate the molecules of 1 x10-g of isopentyl acetate, C, H1402 which are (b) released in a typical bee sting. How many atoms of carbon, hydrogen and oxygen are present in it? Ans. Mass of isopentyl acetate Molar mass of isopentyl acetate (CH1402) = 12x7+14 + 16x2 Number of molecules of isopentyl acetate = Number of molecules of isopentyl acetate = 1x10% Number of molecules of isopentyl acetate 1 molecule of C.H1402 contains 7 atoms of carbon, 14 atoms of hydrogen and 2 atoms of oxygen. Number of carbon atoms = 7 × 4.63 × 1015 Number of carbon atoms = 3.24 × 1016 Number of hydrogen atoms = 14 × 4.63 × 1015 Number of hydrogen atoms = 6.3 × 1016 Number of oxygen atoms = 2 × 4.63 × 1015 Number of oxygen atoms = 9.26 × 1015 Interesting Information! S.Q. What is role of isopentyl acetate in bananas? Ans. Isopentyl acetate (CH1402) is the compound responsible for the scent of bananas. Interestingly, bees release about 1 ug (1 x 10°g) of this compound when they sting. The resulting scent attracts other bees to join the attack: MOLAR VOLUME Define molar volume. Give its relation with mole and Avogadro's number. How

Molar Volume Definition: The volume of one mole of an ideal gas at STP (Standard temperature and pressure) is called molar volume. Value: Its value is equal to 22.414 dm. The value of molar volume is commonly rounded to 22.4 dm? It is denoted by Vm. Example 22.4 dm of CO2 at STP = 44.0g of CO2 = 6:02 x 102 molecules of CO2 = 1 mole of CO2 22.4 dm of H2 gas at STP = 2g of H2 = 6.6.02 x 10^23 molecules of H2 = 1 mole of H2 22.4 dm of NH3 gas at STP = 17g of NH3 = 6.6.02 × 10^23 molecules of NH3 = 1 mole of NH3 Relationship between Molar volume and Avogadro's Number According to Avogadro's law, "Equal volumes of all ideal gases at the same temperature and pressure contain equal numbers of molecules". • This statement is indirectly the same when we say that one mole of an ideal gas at 273.16 K and one atm pressure has a volume of 22.414 dm = 1x10°g = 84 + 14 + 32 = 130g/mol mass -xNA molar mass - × 6.6.02 × 10^23 130 = 4.63 × 10'5 molecules Answer Answer Answer • One mole of a gas has Avogadro's number of particles. • 22.414 dm of various ideal gases at STP will have Avogadro's number of molecules i.e., 6.6.02 × 10^23 • 22.4dm? of a gas at STP is equal to molar mass of a gas its number of moles will be i. Calculation of Volume of gas from Vm If the number of moles of a gas is known, one can calculate it's volume by multiplying number of moles of the gas with molar volume. Volume of a gas = Number of moles × Molar volume V = n x Vm

Density of Gases Definition: Density is defined as the mass per unit volume of a substance. Density = Relationship between density and molar mass • Molar mass of all the gases occupies same volume at STP. • Density of a gas depends on molar mass. • A gas having higher molar mass will have higher density and vice verse. • If the density of gas at STP is determined, its molar mass can be calculated. Quick Check 4.3 Calculate the molar mass of a gas which has density of 1.34g/dm' at STP. Ans. Molar mass of gas = ? Density of gas, = d = 1.34 g/dm Volume of gas at STP = 22.4 dm M= d x v M = 1.34g / dm? × 22.4 dm? Answer M = 30.016 g/mol Molar concentration and relation with moles Definition: Molar concentration of solution is given as mol/dm? moles of a substance (reactant or product) dissolved per volume of a solution in dm Formula: The relationship between number of moles and molar concentration is given by n = C x V C = molar concentration V = volume of the solution Molar Concentration = Mass Volume V , which is the number of C=- V Number of moles Volume in dm Example: Concentration of 0.2 mole of a substance per dm? is: 0.2 mol : = 0.2 mol dm 1dm Mass Molar mass Number of moles No. of particles AvogadroEs no Quick Check 4.4 Calculate the molar concentration of a solution containing 7.9 g of KMnO4 dissolved in 1 dm' of the given solution. The molar mass of KMnO4 is 158g mol. Ans. Mass of KMnO4 = 7.9g Volume of solution = 1dm? Molar concentration = ? Solution: 7.9g n= 158g / mol = 0.05mol No.of moles Molar concentration =. volume in dm 0.05 mol = - 1 dm" Molar concentration = 0.05 mol / dm? 05. Define stoichiometry. Explain its relationships and steps for stoichiometric calculations? Ans. Stoichiometry Meaning: Stoichiometry is derived from Greek words 'stoicheion' means element and 'metron' means measure. Statement: Stoichiometry is a branch of chemistry which tells us the quantitative relationship between reactants and products in a balanced chemical equation. Volume Molar Volume Conc. × Volume Answer Assumptions: (i) Law of conservation of mass and law of definite proportions are obeyed. (ii) No side reaction occurs (wii) All the reactants should be converted into products. Laws followed in stoichiometry • Law of conservation of mass: According to this law "matter (mass) can neither be created nor be destroyed". It's states in terms of stoichiometry, the total mass of reactants should be equal to the total mass of products in a balanced chemical equation. • Law of definite proportions: According to this law a pure compound always contains the same element combined in the same ratio by mass. The balanced chemical equation has the same number of atoms of each element on both sides of equation. It has definite ratio of reactants. and products just as compounds have definite ratios of elements. The ratio is used to calculate the mass or mole of other substances. Explanation: The knowledge of mole, Avogadro's number, molar mass, molar volume and molar concentration, can give quantitative relationships between reactants and products using the balanced chemical equations. Stoichiometric Relationships The following types of relationship can be studied with the help of a balanced chemical equation involving quantities of reactants and products. (i) Mole-Mole Relationship (ii) Mass-Mass Relationship (ill) Volume-Volume Relationship (iv) Mole-Mass Relationship free i'm. (v) Mole-Volume Relationship (vi) Mass- Volume Relationship Example of Stoichiometric calculations: To understand these relationships, we need to interpret information hidden in a balanced chemical equation which is used to make stoichiometric calculations. N2(8 +3H2(g) 2NH 3(g) This equation can be described in different ways: • Mole: 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3. • Number of molecules: 1 molecule of N2 reacts with 3 molecules of H2 to form 2 molecules of NH3. Volume: 22.4 dm? of N2 reacts with 67.2 dm? of H2 to form 44.8 dm of NH3. • Mass: 28.0g of N2 react with 6g of H2 to form 34.0g of NH3. Following are the approaches in Stoichiometric calculations: Keep in Mind! S.Q. What are assumptions of Stoichiometry? Ans. The following assumptions must be made while performing stoichiometric calculations: 1) All the reactants are completely converted into the products. (ii) Law of conservation of mass and law of definite proportions are obeyed (wii) No side reaction occurs. Approach to do Stoichiometric Calculations Given amounts (Step 1): Mass of known solid or volume of known gas, or molar concentration of known solution. Number of moles (Step 2): Calculate number of moles of known solid or volume of a known gas, or molar concentration of known solution using the relevant formula Calculation of Ratio (Step 3): Find the ratio of the known and the unknown reactant or product from the balanced chemical equation. Moles of unknown substance (Step 4): Calculate the number of moles of the unknown reactant or product using the relevant formula. Calculation of unknown amounts (Step 5): Convert the number of moles of the unknown to mass, volume or concentration of the substance. Sample Problem 4.11 (Mole - Mole Conversion) When 3.3 mol of nitrogen reacts with hydrogen to form ammonia, how many moles of hydrogen are consumed in the process? The equation for this reaction is N2(g) + 3H2(g) 2N 3(8) Solution Number of moles of N2 = 3.3 mol Number of moles of H2 =? 1 mole of N2 needs H2 to produce NH3 = 3 mol 3.6 moles of N2 needs H2 to produce NH3 = 3 × 3.3 = 9.9 mol Quick Check 4.5 How many moles of carbon dioxide are produced when 2.25 moles of glucose are used by a person? The oxygen is in excess. The equation for the reaction is: CH,, 06(6) + 602(g) 6CO2(g) + 6H, 0(e) Ans. From balanced chemical equation 1 mole of C6H1206 = 6 moles of CO2 2.25 moles of C6H1206 = 6 × 2.25 moles of CO2 = 13.5 moles of CO2 13.5 moles of CO2 are produced when 2.25 moles of glucose are used by a person. Sample Problem 4.12 (Mass-Mass Conversion) Calculate the mass of Al needed to react completely with 32.0g of iron (III) oxide according to the equation given below: 2Alg) + Fe,O3(s) Solution Molar mass of Fe2O3, M = 159.6 g/mol Number of moles of Fe203 m n M 159.6 g/mol →→ Al, O3(s) + 2 Fe(s) 32.0g - = 0.02 mol From the balanced equation, 1 mol of Fe2O3 reacts with 2 moles of Al, therefore, number of moles of Al that reacts with 0.02 mole of Fe203=2 x 0.02 = 0.04 mol. Mass of Al=n X M = 0.04 mol × 27g mol Mass of Al = 1.08 g Quick Check 4.6 FezO3, an ore of iron is called Hematite. CO can reduce it to get free Fe as below: (g) - Fe, 0 3(s) + 3CO How much Fe can be produced from 160g of FezOз? Ans. Mass of Fe203 = 160g Molar mass of Fez O3 = 159.59g/mol 160 - = 1.0019mol Moles of Fez O3 = - 109.0 From balanced chemical equation: (g) Fe, O 3(s) + 3CO → 2Fe(s) + 3CO2(g) Fe Fez03 1 2 1.0019 2 × 1.0019 2.0038 moles Mass of Fe = moles x moles mass of Fe = 2.0038 × 56 Mass of Fe = 112.213g Answer Sample Problem 4.13(Volume-Volume) Calculate volume of ammonia that can be produced by the reaction of 100 dm? of hydrogen with excess of nitrogen at STP. The balanced chemical equation for the reaction is: N2(g) + 3H2(g) - Solution Volume of hydrogen Volume of ammonia 67.2 dm? (3 moles) of H2 produce ammonia = 44.8 dm3(2 mol) 1 dm' of H2 produce ammonia 100 dm of H2 produce ammonia 1506.7 volume of ammonia produced by the reaction of 100 dm of H2 with excess nitrogen Quick Check 4.7 Calculate the volume of carbon dioxide produced at STP when 4.5 dm of methane is burnt by a person. The oxygen is in excess. The equation for the reaction is: CH 4(g) + 202(g)CO 2(g) + 2H,°(g) Ans. From balanced chemical equation 2(g) CH4(g) +20 →CO2(g) + 2H,°(g) CO2 CH4 22.4dm' 22.4dm3 4.5dm' 4.5 dm? → 2Fe() + 3CO2(g) →→2 NH 3(g) = 100 dm =? 2 44.8 = = 3 67.2 = 2x100 = 66.7dm° Equal moles of any gases at STP have equal volumes. So, the volume of CO2 produced is 4.5dm' that is same volume of methane because they have equal number of moles. Sample Problem 4.14 (Mole-Mass Calculations) Solid lithium hydroxide LiOH is used in space vehicles. It is employed to remove exhaled carbon dioxide from the living environment by forming solid lithium carbonate and liquid water. Calculate the mass of LizCO3 that can be produced by 20.0 mol of LiOH. 2 LiOH g) + CO2(g) Li, CO 3(s) + H2O) Solution According to the given balanced chemical equation, 2 moles of LIOH produces 20.0 moles of LiOH produces Mass of LizCO3 produced Mass of LizCO3 produced Thus. 739.0 g LizCO3 will be produced from 20.0 moles of LiOH. Quick Check 4.8 Calculate the mass of sodium hypochlorite (NaOCl), a household bleach, produced by the reaction of 2.25 moles of chlorine with excess sodium hydroxide. The balanced equation is 2NaOH (q) + Cl 2(g) NaOCl (ag) + NaCl (ag) + H2O (e) Ans. 2NaOH (ag) + Cl - NaOC(ag) + NaCl (ag) + HI, 0(e) From balanced chemical equation Cl2 NaOCl 2.25 2.25 So 2.25 moles of NaOCl are produced by the reaction of 2,25 moles of Cl with excess of NaOH Mass of NaOCl = 2.25 × 74.5 Mass of NaOCl = 167.625g Sample Problem 4.15 (Mass-mole calculations) Baking soda (NaHCO3) acts as an antacid. It can neutralize excess hydrochloric acid (HCl) secreted by the stomach according to equation. NaHCO 3(g) + HC (ag) NaC l (aq) + HO (c) + CO 2(a) How many moles of HCl will be neutralized by 2.1 g of baking soda? Solution: Molar mass of NaHCO3 = 84.0g/mol 2.1g Moles of NaHCO3 = 84.0 g/mol = 0.025 mol Stoichiometrically, the mole ratio of HCl and NaHCO3 is 1: 1. Hence moles of HCl used = 0.025 mol Thus 2.1g of NaHCO3 will neutralize 0.025 moles of HCl. = 1 mol Li, CO3 ==×20.0 = 10.0 molLi, CO, = No.of mole Molar mass = 10.0 mol× 73.9 g/mol-1 = 739.0 g Sample Problem 4.16 (Mass-Volume Conversion) What volume of hydrogen at STP will be produced when 7.0g of iron are reacted with an excess of sulfuric acid? Fe(s) + H,SO A(ag) Solution: Molar mass of Fe (M) = 55.8g/ mol m =- Number of moles of iron (n) M 7.0 g =- 55.8 g/mol = 0.125 mol From the balanced equation, 1 mol of iron produces 1 mole of hydrogen. So, number of moles of H2 = 0.125 mol Volume of H2 in dm3' = molar volume X moles of H2 = 22.4 dm' mol-1× 0.125 mol = 2.8 dm Volume of H2 in dm 06. Define limiting reactant with example. Give strategy for its identification. Ans. Limiting Reactant Definition: The reactant which controls the amounts of products formed in a chemical reaction and is consumed earlier is called the limiting reactant or reagent. The maximum amount of the product formed depends upon the amount of limiting reactant in the reaction mixture. Example: A large quantity of oxygen in a chemical reaction makes things burn more rapidly. In this way, excess of oxygen is left behind at the end of reaction ant the other reactant, i.e. fuel, is consumed earlier. This reactant which is consumed earlier is called the limiting reactant. The amount of product that forms is limited by the reactant that is completely used. Once this reactant is consumed, the reaction stops and no additional product is formed. Strategy for the identification of limiting reactant To identify a limiting reactant, the following three steps are performed. (i) Calculate the number of moles from the given amounts of reactants. (ii) Find out the number of moles of product with the help of a balanced chemical equation. (wii) Identify the reactant which produces the least amount of product as limiting reactant and the other as an excess reactant Did You Know! S.Q. Give role of excess reactant in combustion. Ans. Fire is a combustion reaction in which fuel and oxygen, O2, combine, usually at high temperatures, to form water and carbon dioxide. Once the fire has started, it is self- supporting. An effective way to quench a fire is smothering, which reduces the amount of available oxygen below the level needed to support combustion. In other words, smothering decreases the amount of excess reactant, foams, inert gas, and CO2 are effective substances for smothering. Sample Problem 4.17 (Limiting Reactant) Calculate the mass of N2 produced from 1.81g of NH3 (molar mass = 17.0g mol-1) and 90.4g of CuO (molar mass = 79.5g/mol ') according to following balanced equation: 2NH 3(8) +3 CuO Solution: = Moles of NH3 18.18 of NH3 = 1.06 mol 17.0 g mol Moles of CuO = 90.4 gof Cuo - = 1.14 mol 79.5 g mol In balanced equation, Cuo N2 1 • • 3 1.14 - x1.14 = 0.38 mol NH3 : • • 1 2 1 ple 51d 1.06 -x 1.06=0.53 mol i'm. Thus, CuO is the limiting reactant and the number of moles of N2 produced will be 0.38 mol. Hence, mass of N2 produced = n X M = 0.38 mol × 17.0g mol-1 Mass of N2 = 6.46g Sample Problem 4.18 (Limiting Reactant) When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates down. Calculate the mass of PbSO4 formed when 1.25 dm of 0.05 mol dm* Pb(NO3)2 and 2.00 dm of 0.025 mol dm Na2SO4 are mixed Na,SO 4(a) + Pb (NO3 2(aq) PbSO 4(8) + 2 NaNO 3(ag) Solution: The net ionic equation is Pb2+ Since 0.05 mol dm3 Pb(NO3)2 contains 0.05 mol dm3 Pb2t ions. No. of moles = Concentration (mol dm3) × Volume (dm) n= CV moles of Pb?t ions = 0.05 mol dm3 × 1.25 dm = 0.0625 mol moles of SO4? ions = 0.025 mol dm " ×2.00 dm = 0.05 mol As Piźt and SO4? react in a 1: 1 ratio, here, SO4? (0.05 mol) will be consumed earlier than pb?+ (0.0625 mol). The amount of SO? will be limiting. The reason is that 0.05 mole of SO* is less than 0.0625 mole of Pb?t. Since the Pbt ions are present in excess, only 0.05 mole of solid PbSO4 will be formed. The mass of PbSO4 formed can be calculated using the molar mass of PbSO4 (303.3g/mol-1): Mass of PbSO4 = 0.05 mol × 303.3g/mol-1 = 15.2g

Excess Reactant or non-limiting reactant Definition: The reactant which are in larger amounts (according to stoichiometry of reaction) and remain unreached at the end of the reaction are called "excess reagents" (or excess reactants). Explanation: In many chemical processes, the quantities of the reactants are usually not present in the proportions indicated by the balanced chemical equation. Frequently, a large amount of inexpensive reactant is supplied. Example Consider the reaction between hydrogen and oxygen to form water. 2H 2(g) + 02(g) → → 2H20(1) • When we take 2 moles of hydrogen (4g) and allow it to react with 2 moles of oxygen (64g), then we will get only 2 moles (36g) of water. Actually, we will get 2 moles (36g) of water because 2 moles (4g) of hydrogen react with 1 mole (32g) of oxygen according to the balanced equation. Since less hydrogen is present as compared to oxygen, so hydrogen is a limiting reactant. • When 1 mole of 02 and 1 mole of H2 are mixed, all the H2 will react completely and O2 will be left unreached because for 1 mole of H2, ½ mole of O2 is required. The remaining 1 - mole will be excess. Importance of non-limiting reactant Frequently a large amount of inexpensive reactant is supplied because of the following reasons. • To produce maximum amount of product To increase the rate of reaction • To ensure that whole of the mass of expensive reactant is completely converted to the desired product. Sample Problem 4.19 (Excess Reactant) Natural gas consists primarily of methane (CH4). The complete combustion of methane (CH4) gives carbon dioxide (CO2) and water. 2(t2H,(g) CHA(8) +202(g) (a) How many grams of CO2 can be produced when 30g of CH4 and 50g of O2 are allowed to combine? (b) How many grams of excess reagent are left unreached after the completion of reaction? Solution: (a) Step 1: Write balanced chemical equation. Step 2: Convert the given mass of both the reactants into their moles. Moles of CH, = Biven mass of CHA = - molar mass of CH, 16 g/mol given mass of 02=- Moles of 0, = molar mass of O2 Step 3: Calculate the number of moles of product from each reactant. Compare the number of moles of CH4 with those of CO2. From the balanced chemical equation. 1 mol of methane produces CO2 = 1 mol 1.875 mol of methane produces CO2 = 1 x 1.875 mol = 1.875 mol of CO2 Compare the number of moles of 02 with those of CO2. From the balanced chemical equation, we know: 2 mol of oxygen produces COz = 1 mol 1.563 mol of oxygen produce CO2 = 0.5 x 1.563 mol = 0.7815 mol of CO2 From the above calculation, it is clear that the limiting reactant is O2 because it produces lesser amount (moles) of product (CO2) than CH4. Step 4: Convert the moles of the product into mass. Mass of CO2 in grams = Moles of CO2 X Molar mass of CO2 ale G0 = 0.7815 mol × 44g mol-1 = 34.39g Step 5: The quantity of limiting reactant can also be used to calculate the quantity of excess reactant used 2 mol of O2 reacts with moles of CH4 = 1 mol 1 1.563 mol of 02 reacts with mol of CH4= - x1.563 mol = 0.7815 mol Step 6: The mass of methane (excess reagent) is equal to the starting quantity minus the amount used during the reaction. Number of moles of CH4 in excess = Quantity taken - Quantity used = 1.875 mol - 0.7815 mol = 1.0935 mol Excess mass of CH4 (excess reagent) = 1.0935 X 16.0 = 17.5g Quick Check 4.9 Which of the following reaction mixtures could produce the greatest amount of product when they combine according to the reaction given below? 12(g) +312(g) - (a) 1 mole of N2 and 3 moles of H2 (b) 2 moles of N2 and 3 moles of H2 (c) 1 mole of N2 and 5 moles of H2 (d) 3 moles of N2 and 3 moles of H2 (e) Each produces the same amount of product Ans. (a) Produces 2 moles of NH3 (b) Produces 2 moles of NH3 because 1 mole of N2 remains unreached 30 g - = 1.875 mol 50 g - = 1.563 mol 32 g/mol →2N 3(8) (c) Produces 2 moles of NH3 because 2 moles of H2 remains unreached (d) Produces 2 moles of NH; because 2 moles of N2 remain unreached (e) Is the correct answer, because all conditions produce same moles of NH3.

Actual and Theoretical yield Actual yield Definition: The amount of the products obtained in a chemical reaction is called the actual yield of that reaction. Theoretical yield Definition: The amount of the products calculated from the balanced chemical equation represents the theoretical yield Explanation: The theoretical yield is the maximum amount of the product that can be produced by a given amount of a reactant, according to balanced chemical equation. Theoretical yield is always greater than actual yield. In most chemical reactions the amount of the product obtained is less than the theoretical yield Reason: There are following reasons for that: • Physical Process: The process like filtration, separation by distillation, separation by a separating funnel, washing, drying and crystallization, if not properly carried out, decrease the actual yield. • Side Reaction: Some of the reactants might take part in a competing side reaction and reduce the amount of the desired product. So, in most of the reactions the actual yield is less than the theoretical yield • Reversible Reaction: A reaction may be reversible. Therefore, the amount of the product will be reduced by the backward reaction. Efficiency of reaction A chemist is usually interested in the efficiency of a reaction. The efficiency of a reaction is expressed by comparing the actual and theoretical yields in the form of percentage (%) yield. % Yield = Acuital yield Theoretical yield Greater the %age yield, higher will be the efficiency of reaction and vice verse. Quick Check 4.10 When limestone (CaCO3) is roasted, quicklime (CaO) is produced according to the following equation. CaCO 3(s) The actual yield of CaO is 2.5kg, when 4.5kg of lime stone is roasted. What is the percentage yield of this reaction? Ans. Given Data = 4.5kg = 4500g Mass of CaCO3 = 2.5kg = 2500g Actual yield Molar mass of CaCO3 = 100g/mol Molar mass of CaO = 56g/mol -×100 2(g) → Cao (s) + CO Solution: From the balanced chemical equation. CaO CaCO: 56g 100g 56g 1g 100- - ×4500 = 2520 g 4500 g 100 So Theoretical yield = 2520 2500 -x 100 % yield 2520 = 99.2% % yield

Importance of stoichiometry in production and dosage of medicine In the preparation of required dose of a medicine, the optimum amount of the active ingredient in a medicine is, essential to produce desired effects in the patient. Stoichiometry ensures the following things: • The accuracy of drug synthesis. • Any deviation can result in incomplete reaction or contamination with un-reacted reactants or by-products. • Allows chemists to precisely control chemical reactions to produce drugs. • To ensure its efficiency, effectiveness and safe use of drugs. Significance of Stoichiometry in Medicine Stoichiometry is very important in the field of medicine and is used: (i) Actual product: In the preparation of antibiotics, the stoichiometry ensures that each dose matches the active ingredient and target bacteria: (ii) Analysis of cholesterol: To determine the cholesterol level in the blood of patients. Cholesterol is a form of fat that is not all bad. However, cholesterol can have harmful effects. Insulin level: To determine the glucose level in the blood of diabetic patient. Use of (wii) insulin relies on the stoichiometry to precise control of blood sugar levels. Steroids: To determine the steroid and other stimulants in the urine of athletes. Athletes (iv) use steroids and other stimulants to enhance performance and increase strength. (v) Antigens: To determine the concentration of viral antigens in the preparation of vaccine for effective results. (vi) Amount of drug: To determine the amount and number of drugs to give a dosage to a patient. The medicine has no effect when given in small amounts and can cause toxic state or death when given in large amounts. Paracetamol is used as a pain killer and to decrease fever. An overdose may result a blood thinning, organ damage and severe liver damage. SAMPLE PROBLEMS Sample Problem 4.1 Calculate the number of moles present in 20g of NaOH. Solution: Given mass Number of moles = Molar mass - = 0.5 mol n = 20 40 Sample Problem 4.2 Calculate the mass of 0.5 moles of Hcl Solution: Mass of Hcl = Number of moles x Molar mass Mass of Hcl = 0.5 x 36.5 = 18.3 mol Sample Problem 4.3 Calculate the mass of 10-3 mol of MgSO4. Solution: Molar mass of MgSO4 =24 + 96 = 120g/mol-1 Number of moles of MgSO4 = 10-3 Mass of MgSO4 = 103 mol × 120g/mol = 120 x 10=0.12g Sample Problem 4.4 See at page number 109 Sample Problem 4.5 See at page number 109 Sample Problem 4.0 See at page number 109 Sample Problem 4.7 Determine the volume of 2.5 moles of chlorine molecules at STP. Solution: The formula for volume determination at STP is, V= nx Vm Volume of 2.5 mole of Cl2 = 22.4 dm × 2.5 = 56.0 dm Sample Problem 4.8 What is the volume in dm of 4.75 mol of methane (CH4) gas at STP? Solution: The formula for volume determination at STP is, V=nx Vm Volume of methane in dm at STP= 4.75 × 22.4 = 106.4dm Sample Problem 4.9 Calculate the molar mass of a gas which has density of 1.97g/dm? at STP. Solution: Mass of gas at STP = 1.97 × 22.4 = 44.1g mol-1 Sample Problem 4.10 Calculate the molar concentration of a substance containing 27.64g of KzCO3 dissolved in 1 dm of the given solution Solution: Mass of K2CO3 = 27.64g Molar mass of K2CO3 = 138.2 g/mol Given mass Number of moles = Molar mass m n= M 27.64 n= - = 0.2moles 138.2 Volume of solution = 1 dm? Molar Concentration = Number of moles Volume in dm C = 0.2 mol = 0.2 moldm3 1 dm Sample Problem 111 Mole Mole Conversion) See at page number 114 Sample Problem 4.12 (Mass-Mass Conversion) See at page number 114 Sample Problem 4.13(Volume-Volume) See at page number 115 Sample Problem 4.14 (Mole-Mass Calculations) See at page number 116 Sample Problem 4.15 (Mass-mole calculations) See at page number 116 Sample Problem 4.16 (Mass- Volume Conversion) See at page number 117 Sample Problem 4.17 (Limiting Reactant) See at page number 118 Sample Problem 4.18 (Limiting Reactant) See at page number 118 Sample Problem 4.19 (Excess Reactant) See at page number 120 Sample Problem 4.20 (%age Yield) Aspirin (CgHsO4) is prepared by heating salicylic acid, C+H6O3 (molar mass 138.12g mol-1) and acetic anhydride C4H6O3 (molar mass 163.93g mol). Calculate the theoretical yield of aspirin, (molar mass 180.16g mol) when 3.00g of salicylic acid is heated with 6.00g of C4H6O3. What is % yield when actual yield is 3.15g? Solution: + С,H, 03 Salicylic acid Acetic anhydride 1 mol 1 mol 138.12 g 163.93 g →CHg 4(s) + CH, COOH (ag) + →→ C,H,0A CH, COOH Aspirin 1 mol 180.16 g 1 mol of salicylic acid produces aspirin = 1 mol Mass of salicylic acid = 3.00 g 3.0 g Number of moles of salicylic acid= - 138.12 gmo): = 0.022 mol 6.00 g Number of moles of acetic anhydride = 163.93g mol = 0.037 mol Here, salicylic acid is limiting reactant while acetic anhydride is an excess reactant. The amount of salicylic acid controls the yield of product i.e., aspirin. 0.022 moles of salicylic acid produces aspirin = 0.022 moles Mass of Aspirin = 0.022 moles × 180.16g/mol Theoretical yield = 3.96g Actual yield = 2.85 2.85 - × 100 = 71.97% % age yield = 3.96 free i'm.