Chapter 9: Stoichiometry The Arithmetic ti Of Equations

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1 Chapter 9: Stoichiometry The Arithmetic of Equations Chemical Calculations Limiting Reagent and Percent Yield The Arithmetic ti Of Equations -- The Arithmetic of Equations -- Using Everyday Equations Stoichiometry Part of chemistry that deals with numerical relationships in chemical reactions Calculation of quantities of substances involved in chemical equations Recipes: proportional amounts Interpreting chemical equations Information that can be seen or derived from chemical equations Particles Moles Mass Volume -- The Arithmetic of Equations -- Conservation Always conserved Mass Atoms Sometimes conserved Molecules Formula units Moles Volumes of gases

2 -- The Arithmetic of Equations -- Example For the reaction 2H 2 (g) +O( + O 2 (g) 2H 2 O(g), interpret this reaction in terms of the following three relative quantities. Number of representative particles» 2 molecules of hydrogen gas react with 1 molecule of oxygen gas to produce 2 molecules of water vapor. Number of moles» 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor. Masses of reactants and products -- The Arithmetic of Equations -- Masses of reactants and products» In order to relate masses of reactants and products, you must multiply the number of moles of each reactant and product by its molar mass.» H 2 : 2(1.0079) = g/mol» O 2 : 2(15.999) = g/mol» H 2 O: 2(1.0079) + 1(15.999) = g/mol» Reactants: 2H 2 + O 2 = 2(2.0158) = g» Products: 2H 2 O = 2( ) = g» g = g» The Law of Conservation of Mass is preserved. Chemical Calculations l Chemical calculations For all calculations involving two chemicals, you need to know the BALANCED chemical reaction involved. Mole to mole calculations If you know the number of moles for a compound in one part of an equation, you can determine the number of moles for a different compound in another part of the equation (using molar ratio). Examples For the reaction, 2H 2 (g) + O(g) 2 2H 2 O(g), how many moles of oxygen gas must react with hydrogen gas in order to get 0.85 mol water?» For every two moles of water, there is one mole of oxygen gas.

3 Iron metal reacts with oxygen gas to produce iron (III) oxide. How many moles of oxygen gas are needed to react fully with 4.65 moles of iron metal?» Fe + O 2 Fe 2 O 3 (unbalanced)» 4Fe + 3O 2 2Fe 2 O 3 (balanced)» For every three moles of oxygen gas, there are four moles of iron metal. Mass to mole calculations; mole to mass calculations If you know the mass for a compound in one part of an equation, you can determine the mass for a different compound in another part of the equation by first converting to moles and then back to mass (using molar ratio again). Mass A Moles A Moles B Mass B Examples Hydrochloric acid reacts with iron (III) oxide to produce iron (III) chloride and water. Calculate the number of grams of iron (III) oxide needed to react with excess hydrochloric acid in order to produce 468gwater 4.68 water.» HCl + Fe 2 O 3 FeCl 3 + H 2 O (unbalanced)» 6HCl + Fe 2 O 3 2FeCl 3 +3H 2 O (balanced)» Since mass is involved with the two chemical described, find the molar mass of the involved chemicals. H 2 O: 2(1.0079) = g/mol Fe 2O 3 : 2(55.847) + 3(15.999) = g/mol» Like the calculations you did in Chapter 7, always convert to moles first. For every three moles of water, there is one mole of iron (III) oxide. Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. Calculate the number of grams of tin needed to completely react with 10.2 grams of hydrofluoric acid.» Sn + HF SnF 2 +H 2 (unbalanced)» Sn + 2HF SnF 2 + H 2 (balanced)» Since mass is involved with the two chemicals described, d find the molar mass of the involved chemicals. Sn: g/mol HF: 1(1.0079) + 1(18.998) = g/mol» Like the calculations you did in Chapter 7, always convert to moles first. For every one mole of tin, there are two moles of hydrofluoric acid.

4 Other calculations ALWAYS convert to moles first (if not already in moles). If mass is involved, use molar mass. If volume at STP is involved, use molar volume (22.4 L/mol). If representative particles are involved (i.e., atoms, ions, formula units, molecules), use Avogadro s number ( particles/mol). Units A Moles A Moles B Units B Examples Tin reacts with hydrofluoric acid to produce tin (II) fluoride and hydrogen gas. How many grams of SnF 2 can be made by reacting molecules of hydrofluoric acid with excess tin?» Sn + HF SnF 2 + H 2 (unbalanced)» Sn + 2HF SnF 2 + H 2 (balanced)» Since mass of SnF 2 is involved, find the molar mass of SnF 2. Sn: 1(118.69) + 2(18.998) = g/mol» Since molecules (representative particles) are involved for HF, Avogardo s number will be used ( molecules HF/mol HF).» For every one mole of SnF 2, there are two moles of hydrofluoric acid.» Since volume is involved for O 2, molar volume will be used (22.4 L O 2 /mol O 2 ).» For every one mole of O 2, there are two moles of NO. Nitrogen monoxide gas and oxygen gas combine to form the brown gas nitrogen dioxide. How many liters of oxygen gas are required to react fully with 7.8 grams of nitrogen monoxide gas at standard temperature and pressure?» NO + O 2 NO 2 (unbalanced)» 2NO + O 2 2NO 2 (balanced)» Since mass of NO is involved, find the molar mass of NO. NO: 1(14.007) + 1(15.999) = g/mol

5 Limiting iti Reagent And Percent Yield Limiting reagent Any reactant that is used up first in a chemical reaction Determines the amount of product that can be formed in the reaction Excess reagent Any reactant that remains after the limiting reagent is used up in a chemical reaction Reagent present in a quantity that is more than sufficient to react with a limiting reagent Hints If you have more than you calculate you need, that is the excess reagent. If you calculate you need more than you have, that is the limiting reagent. Examples Water can be prepared by the reaction of hydrogen gas with chlorine gas as shown in the equation: 2H 2 (g) + O 2 (g) 2H 2 O(l). Suppose that 6.70 moles of hydrogen gas react with 3.20 moles of oxygen gas. What is the limiting reagent, and how many moles of water are produced? First, calculate how many moles of each are needed compared to the other chemical. Second, once you have determined which is the limiting reagent, use the amount of limiting reagent to figure out the other amounts of chemicals as asked. For every one mole of O 2, there are two moles of H 2. Amount of hydrogen gas needed for the shown amount of oxygen gas: amount of oxygen gas: Since the calculation shows you need 6.40 mol H 2 but you have 6.70 mol H 2 (more than you need), H 2 must be the excess reagent, meaning O 2 is the limiting reagent. You could begin with the amount of H 2 instead and figure out the same answer. Since the calculation shows you need 3.35 mol O 2 but you have only 3.20 mol O 2 (less than you need), H 2 must be the excess reagent, meaning O 2 is the limiting reagent. (You can calculate either way.)

6 Since oxygen gas is the limiting reagent, any calculations predicting other chemical reactions will involve the 3.20 mol O 2. Also, you know that there are 2 moles of H 2 O for every 1 mole of O 2. Aluminum metal reacts with aqueous copper (II) sulfate to produce copper metal and aluminum sulfate grams of aluminum react with 13.5 grams of copper (II) sulfate. (A) What is the limiting reagent? (B) How many grams of excess reagent are there? (C) What is the maximum number of grams of aluminum sulfate that can be produced? Al(s) + CuSO 4 (aq) Al 2 (SO 4 ) 3 (aq) + Cu(s) (Unbalanced) 2Al(s) + 3CuSO 4 (aq) Al 2 (SO 4 ) 3 (aq) + 3Cu(s) (Balanced) First, calculate how many grams of each are needed compared to the other chemical. You will need the molar mass of each chemical to do this. Al: g/mol CuSO 4 : 1(63.546) + 1(32.06) + 4(15.999) = g/mol Amount of aluminum needed for the shown amount of copper (II) sulfate: To figure out the amount of excess reagent, take the amount you have of aluminum and minus how much you need g Al 1.52 g Al = 1.28 g Al Since the calculation shows you need 1.52 g Al but you have 2.80 g Al (more than you need), Al must be the excess reagent, meaning CuSO 4 is the limiting reagent. (You could begin with the amount of CuSO 4 instead and figure out the same answer.) Since copper (II) sulfate is the limiting reagent, any calculations predicting other chemical reactions will involve the 13.5 g CuSO 4. There is 1 mole of Al 2 (SO 4 ) 3 for every 3 moles of CuSO 4. Molar mass of Al 2 (SO 4 ) 3 : 2(26.982) + 3(32.06) + 12(15.999) = g/mol

7 (A) What is the limiting reagent? CuSO 4 (B) How many ygrams of excess reagent are there? 1.28 g Al (C) What is the maximum number of grams of aluminum sulfate that t can be produced? d? 9.65 g Al 2 (SO 4 ) 3 For the reaction N 2 + 3H 2 2NH 3, you are provided 15.2 g of nitrogen gas and 16.8 g of hydrogen gas to react to produce ammonia (NH 3 ). (A) What is the limiting reagent? (B) How many grams of excess reagent are there? (C) What is the maximum number of grams of ammonia that can be produced? First, calculate how many grams of each are needed compared to the other chemical. You will need the molar mass of each chemical to do this. H 2 : 2(1.0079) = g/mol N 2 : 2(14.007) = g/mol Amount of hydrogen gas needed for the shown amount of nitrogen gas: amount of nitrogen gas: Since the calculation shows you need 3.28 g H 2 but you have 16.8 g H 2 (more than you need), H 2 must be the excess reagent, meaning N 2 is the limiting reagent. (You could begin with the amount of H 2 instead and figure out the same answer.) 16.8 g H g H 2 = g H 2 = 13.5 g H 2 Since nitrogen gas is the limiting iti reagent, any calculations predicting other chemical reactions will involve the 15.2 g N 2. There are 2 moles of NH 3 for every 1 mole of N 2. Molar mass of NH 3 : 1(14.007) + 3(1.0079) = = g/mol To figure out the amount of excess reagent, take the amount you have of hydrogen gas and minus how much you need.

8 (A) (B) (C) What is the limiting reagent? N 2 How many grams of excess reagent are there? 13.5 g H 2 What is the maximum number of grams of ammonia that can be produced? 18.5 g NH 3 Theoretical yield Amount of product that could theoretically form during a reaction Calculated based on the limiting reagent and represents the maximum amount of product formed from the limiting reagent Actual yield Amount of product that forms when a reaction is carried out in a laboratory Amount usually given in a problem as actually all being formed Percent yield Ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage Serves as a measure of the efficiency of a reaction Formula for percent yield or