Chapter 4 Chemical Quantities and Aqueous Reactions
Stoichiometry The study of the numerical relationship between chemical quantities in a chemical reaction
Making Pizza The number of pizzas you can make depends on the amount of the ingredients you use. 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza This relationship can be expressed mathematically 1 crust : 5 oz. sauce : 2 cu cheese : 1 pizza If you want to make more than one pizza, you can use the amount of cheese you have to determine the number of pizzas you can make.
Predicting Amounts from Stoichiometry The amounts of any other substance produced or consumed in a chemical reaction can be determined from the amount of just one substance.
According to the following equation, how many moles of water are made in the combustion of 0.10 moles of glucose? C6H12O6 + 6 O2 6 CO2 + 6 H2O glucose + oxygen gas carbon dioxide + water 1 mol glucose 6 mol water conversion factors 6 mol water 1 mol glucose 1 mol C6H12O6 6 mol H2O conversion factors 6 mol H2O 1 mol C6H12O6 mol C 6 H 12 O 6 mol H 2 O 0.10 mol C6H12O6 x 6 mol H2O = 0.60 mol H2O 1 mol C6H12O6
The amounts of any other substance produced or consumed in a chemical reaction can be determined from the amount of just one substance. Moles of A Coefficients Moles of B
Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 10 15 g of octane (C8H18). C8H18(l) + O2(g) CO2(g) + H2O(g) 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) g C 8 H 18 mol C 8 H 18 mol CO 2 g CO 2 2 mol C8H18 16 mol CO2 conversion factors 16 mol CO2 2 mol C8H18 1 mol C8H18 114.22 g C8H18 conversion factors 114.22 g C8H18 1 mol C8H18 1 mol CO2 44.01 g CO2 conversion factors 44.01 g CO2 1 mol CO2
Estimate the mass of CO2 produced in 2007 by the combustion of 3.5 x 10 15 g of octane (C8H18). 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) g C 8 H 18 mol C 8 H 18 mol CO 2 g CO 2 2 mol C8H18 16 mol CO2 conversion factors 16 mol CO2 2 mol C8H18 1 mol C8H18 114.22 g C8H18 1 mol CO2 44.01 g CO2 conversion factors conversion factors 114.22 g C8H18 1 mol C8H18 44.01 g CO2 1 mol CO2 3.5 x 10 15 g C8H18 x 1 mol C8H18 16 mol CO2 x x 44.01 g CO2 114.22 g C8H18 2 mol C8H18 1 mol CO2 = 1.0789 1.1 x 10x 16 10 16 g CO2
How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis? 6 CO2 + 6 H2O C6H12O6 + 6 O2 g CO 2 mol CO 2 mol C 6 H 12 O 6 g C 6 H 12 O 6 1 mol CO2 44.01 g CO2 conversion factors 44.01 g CO2 1 mol CO2 1 mol C6H12O 6 mol CO2 conversion factors 6 mol CO2 1 mol C6H12O6 1 mol C6H12O6 180.2 g C6H12O6 conversion factors 180.2 g C6H12O6 1 mol C6H12O6 37.8 g CO2 x 1 mol CO2 x 1 mol C6H12O x 180.2 g C6H12O6 44.01 g CO2 6 mol CO2 1 mol C6H12O6 = 25.796 25.8 g C6H12O6
Lead (IV) oxide decomposes to yield lead(ii) oxide and oxygen gas. How many grams of O2 can be made from the decomposition of 100.0 g of PbO2? 2 PbO2(s) 2 PbO(s) + O2(g) (PbO2 = 239.2, O2 = 32.00) g PbO 2 mol PbO 2 mol O 2 g O 2 1 mol PbO2 239.2 g PbO2 1 mol O2 2 mol PbO2 32.00 g O2 1 mol O2 100.0 g 1 mol PbO2 1 mol O2 32.00 g O2 PbO2 x x x 239.2 g PbO2 2 mol PbO2 1 mol O2 = 6.68896 6.689 g O2
Stoichiometry Road Map Grams of A Grams of B Molar Mass Moles of A Mole to Mole Ratio from balanced equation Moles of B Avogadro s Number Particles of A Particles of B
More Making Pizzas 1 crust + 5 oz. tomato sauce + 2 cu cheese 1 pizza What would happen if we had 4 crusts, 15 oz. tomato sauce, and 10 cu cheese? Limiting reagent Theoretical yield
Limiting and Excess Reactants in the Combustion of Methane CH4(g) + O2(g) CO2(g) + H2O(g) CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)
If we have five molecules of CH4 and eight molecules of O2, which is the limiting reactant? 8 mol O2 x = 4 mol of CO2 1 mol CO2 2 mol O2
The Limiting Reactant For reactions with multiple reactants, it is likely that one of the reactants will be completely used before the others. When this reactant is used up, the reaction stops and no more product is made.
How many moles of Si3N4 can be made from 1.20 moles of Si and 1.00 mole of N2 in the reaction: 3 Si + 2 N2 Si3N4? 1.20 mol Si x 1.00 mol Si3N4 = 0.400 mol Si3N4 3.00 mol Si Limiting reactant 1.00 mol Si3N4 1.00 mol N2 x = 0.500 mol Si3N4 2.00 mol N2 Theoretical yield
More Making Pizzas Let s now assume that as we are making pizzas, we burn a pizza, drop one on the floor, or other uncontrollable events happen so that we only make two pizzas. The actual amount of product made in a chemical reaction is called the actual yield. We can determine the efficiency of making pizzas by calculating the percentage of the maximum number of pizzas we actually make. In chemical reactions, we call this the percent yield.
How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? g NH3 mol NH3 mol N2 g mol mol mol smaller amount is from }limiting reactant g CuO mol CuO mol N2
How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? 9.05 g NH3 x 1.00 mol NH3 x 1.00 mol N2 = 0.2657 mol N2 17.03 g NH3 2.00 mol NH3 45.2 g CuO x 1.00 mol CuO x 1.00 mol N2 = 0.1894 mol N2 79.55 g CuO 3.00 mol CuO Limiting reactant Smaller # moles of N2 Theoretical yield
How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are made, what is the percent yield? smaller mol N2 g N2 Theoretical Yield Actual Yield = % Yield Theoretical Yield
How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of copper(ii) oxide? 2 NH3(g) + 3 CuO(s) N2(g) + 3 Cu(s) + 3 H2O(l) If 4.61 g of N2 are isolated, what is the percent yield? 45.2 g CuO x 1.00 mol CuO x 1.00 mol N2 = 0.1894 mol N2 79.55 g CuO 3.00 mol CuO 0.1894 mol N2 x 28.02 g N2 = 5.307 g N2 1.00 mol N2 0.189 mol N2 x 28.02 g N2 = 5.30 g N2 1.00 mol N2 Theoretical yield 4.61 g N2 5.30 g N2 x 100% = 87.0 % 4.61 g N2 5.31 g N2 percent yield x 100% = 86.8 %
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) kg C g C mol C mol Ti k g g mol mol mol smaller amount is from limiting }reactant kg TiO2 g TiO2 mol TiO2 mol Ti
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) smaller mol Ti g Ti kg Ti Theoretical Yield Actual Yield Theoretical Yield = % Yield
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) Collect needed relationships: 1000 g = 1 kg Molar Mass Ti = 47.87 g/mol Molar Mass C = 12.01 g/mol Molar Mass TiO2 = 79.87 g/mol 1 mole TiO2 : 1 mol Ti 2 mole C : 1 mol Ti
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) 1000 g C 1.00 mol C 1.00 mol Ti 28.6 kg C x x x 1 kg C 12.01 g C 2.00 mol C = 1.1907 x x 10 10 3 3 mol Ti 88.2 kg TiO2 x 1000 g TiO2 x 1.00 mol TiO2 x 1.00 mol Ti 1 kg TiO2 79.87 g TiO2 1.00 mol TiO2 = 1.1043 x x 10 10 3 3 mol Ti limiting reactant smallest moles of Ti Theoretical yield
When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the limiting reactant, theoretical yield, and percent yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g) 1.10 x 10 3 mol Ti x 47.87 g Ti x 1.00 kg Ti = 52.9 kg Ti 1 mol Ti 1000 g Ti theoretical yield percent yield