Chapter 3: STOICHIOMETRY: MASS, FORMULAS, AND REACTIONS

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1 Chapter 3: STOICHIOMETRY: MASS, FORMULAS, AND REACTIONS Problems: , 3.11, , , , , 3.134, , THE MOLE Stoichiometry (STOY-key-OM-e-tree): quantitative study of reactants and products in a chemical reaction Interpreting a Chemical Equation H2 (g) + Cl2 (g) 2 HCl (g) 1 molecule 1 molecule 2 molecules It follows that any multiples of these coefficients will be in same ratio! 2 H2 (g) + O2 (g) 2 H2O(g) 1000 molecule(s) molecule(s) molecule(s) N molecule(s) molecule(s) molecule(s) Since N = Avogadro s # = molecules = 1 mole 2 H2 (g) + O2 (g) 2 H2O(g) mole(s) mole(s) mole(s) Thus, the coefficients in a chemical equation give the mole ratios of reactants and products. Consider the following: 2 C2H6 (g) + 7 O2 (g) 4 CO2 (g) + 6 H2O (g) 1. How many moles of O2 will react with 2.50 moles of C2H6? 2. How many moles of CO2 form when 3.50 moles of O2 completely react? CHEM 161: Chapter 3 v0313 page 1 of 6

2 3.5 STOICHIOMETRIC CALCULATIONS AND THE CARBON CYCLE Mass-Mass Stoichiometry Problems MASS OF KNOWN Molar Mass MOLES OF KNOWN MOLE-MOLE Ratio MOLES OF UNKNOWN Molar Mass MASS OF UNKNOWN 3.4 Combustion Reactions: CxHy + O2(g) CO2(g) + H2O(g) CxHyOz + O2(g) CO2(g) + H2O(g) Hydrocarbons (compounds with only C and H) and hydrocarbon derivatives (compounds with only C, H and O) burn in O2 to produce CO2 gas and steam, H2O(g). Ex. 1: Many home barbecues are fueled with propane gas (C3H8). Write the balanced equation for the combustion of propane, then calculate the mass (in kg) of carbon dioxide produced upon complete combustion of liquid propane from a 5.0 gal tank. (Note that 1 lb.=453.6 g, and the density of liquid propane at 60 F is about 4.2 lbs. per gallon.) Ex. 2 Everclear is a brand of grain alcohol that can be as high as 190 proof (or 95% ethanol, C2H5OH, by volume). Calculate the mass of carbon dioxide produced upon complete combustion of the ethanol in a 750 ml bottle of Everclear. Write the balanced chemical equation for the combustion of ethanol. (The density of ethanol is g/ml.) CHEM 161: Chapter 3 v0914 page 2 of 6

3 3.9 LIMITING REACTANTS (or LIMITING REAGENTS) AND PERCENT YIELD In practice, reactants will not always be present in the exact amounts necessary to be converted completely into products. Some reactants (usually the more expensive) are only present in a limited supply, so these are almost always completely used up limiting reactant (or limiting reagent) since it limits the amount of product made Some reactants (usually the less expensive) are present in larger amounts and are never completely used up reactant(s) in excess GUIDELINES for Solving Limiting Reactant Problems 1. Calculate the mass or the # of moles of the 2 nd reactant needed to completely react with the 1 st reactant. If the moles needed is greater than the number of moles present for the 2 nd reactant That 2 nd reactant will run out before the 1 st reactant. The 2 nd reactant = the limiting reactant, and the 1 st reactant is in excess. If the moles needed is less than the number of moles present for the reactant, The 1 st reactant = the limiting reactant, and the 2 nd reactant is in excess. 2. Use the amount of the limiting reactant present to solve for the mass or # of moles of product that can be made. Consider the reaction to produce ammonia: N2(g) + 3 H2(g) 2 NH3(g) Ex. 1. a. If 40.0 g of N2 react with 10.0 g of H2, what mass of ammonia is produced? b. The limiting reactant is, and the reactant in excess is. c. What mass of the reactant in excess remains after the reaction? CHEM 161: Chapter 3 v0313 page 3 of 6

4 CALCULATING PERCENT YIELD: Percent yield = actual yield 100% theoretical yield theoretical yield: Amount of product one should get based on the chemical equation and the amount of reactants present One generally calculates this in grams from info given actual yield: Amount of product one actually obtains Generally smaller than the theoretical yield because of impurities and other adverse conditions in the lab This is generally determined experimentally in the lab or given for a problem in lecture. Ex. 1: a. For the reaction of 40.0 g of N2 with 10.0 g of H2 completed on the previous page, the theoretical yield of ammonia was determined to be what? theoretical yield = (not rounded to s.f. yet) b. If 45.7 g of ammonia were produced, calculate the percent yield for the reaction. percent yield = Ex. 2: Consider the following reaction: 2 KClO3(s) 2 KCl(s) + 3 O2(g) What is the percent yield if 50.0 g of KClO3 decomposes to produce 16.4 g of oxygen gas? CHEM 161: Chapter 3 v0914 page 4 of 6

5 Ex. 3 Consider the following reaction: 2 Na3PO4(aq) + 3 MgSO4(aq) Mg3(PO4)2(s) + 3 Na2SO4(aq) a. What mass of precipitate can be produced when 50.0 g of sodium phosphate react with 50.0 g of magnesium sulfate? Indicate the limiting reactant and the reactant in excess below. mass of precipitate produced = limiting reagent = reactant in excess = b. What mass of the reactant in excess remains after the reaction? c. What is the percent yield if 39.2 g of precipitate are actually produced? d. What ions are present in solution after the reaction described in part a? CHEM 161: Chapter 3 v0313 page 5 of 6

6 Ex. 4 Calculate the mass of methane (CH4) required to produce 10.0 kg of carbon dioxide if the percent yield for the reaction is 88.8%. Ex. 5 Consider the thermal decomposition of N2O5: 2 N2O5(g) 4 NO2(g) + O2(g) If the percent yield for the reaction is 96.8%, and the density of oxygen gas is 1.31 g/l, calculate the mass of N2O5 required to produce 50.0 L of oxygen gas. CHEM 161: Chapter 3 v0914 page 6 of 6