Chapter 6 Chemical Reactions and Quantities

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Chapter 6 Chemical Reactions and Quantities Mass Calculations for Reactions

Guide to Calculating the Masses of Reactants and Products

Example of Converting Moles to Grams Suppose we want to determine the mass (g) of NH3 that can form from 2.50 moles of N2. N2(g) + 3H2(g) 2NH3(g) STEP 1 Given: 2.50 moles of N2 Need: mass (g) of NH3 STEP 2 Plan: moles of N2 moles of NH3 grams of NH3

Moles to Grams STEP 3 Write conversions factors: 1 mole of N2 = 2 moles of NH3 1 mole N2 and 2 moles NH3 2 moles NH3 1 mole N2 1 mole of NH3 = 17.0 g of NH3 1 mole of NH3 and 17.0 g of NH3 17.0 g of NH3 1 mole of NH3 STEP 4 Setup to calculate g of NH3: 2.50 moles N2 x 2 moles NH3 x 17.0 g NH3 1 mole N2 1 mole NH3 = 85.0 g of NH3

Learning Check How many grams of O2 are needed to produce 0.400 mole of Fe2O3 in the following reaction? 4Fe(s) + 3O2(g) 2Fe2O3(s) 1) 38.4 g of O2 2) 19.2 g of O2 3) 1.90 g of O2

Solution 2) 19.2 g of O2 STEP 1 Given: 0.400 moles of Fe2O3 Need: mass (g) of O2 STEP 2 Write a plan to calculate grams of O2: moles of Fe2O3 moles of O2 grams of O2 STEP 3 Write conversion factors: 3 moles of O2 = 2 moles of Fe2O3 3 moles O2 and 2 moles Fe2O3 2 moles Fe2O3 3 moles O2

Solution (continued) STEP 3 (continued) 1 mole of O2 = 32.0 g of O2 1 mole O2 and 32.0 g O2 32.0 g O2 1 mole O2 STEP 4 The setup to calculate g of O2 is 0.400 mole Fe2O3 x 3 moles O2 x 32.0 g O2 = 19.2 g of O2 2 moles Fe2O3 1 mole O2 mole-mole factor molar mass

Learning Check 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 1) 88.6 g of C2H2 Acetylene gas, C2H2, burns in the oxyacetylene torch for welding. How many grams of C2H2 are burned if the reaction produces 75.0 g of CO2? 2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(g) 1) 88.6 g of C2H2 2) 44.3 g of C2H2 3) 22.2 g of C2H2

Solution g of CO2 moles of moles of g of C2H2 CO2 C2H2 STEP 1 Given: 75.0 g of CO2 Need: mass (g) of C2H2 STEP 2 Write a plan to calculate grams of C2H2: g of CO2 moles of moles of g of C2H2 CO2 C2H2 STEP 3 Write conversion factors: 1 mole of CO2 = 44.0 g of CO2 1 mole CO2 and 44.0 g CO2 44.0 g CO2 1 mole CO2

Solution (continued) STEP 3 Write conversion factors (continued): 4 moles of CO2 = 2 moles of C2H2 4 moles CO2 and 2 moles C2H2 2 moles C2H2 4 moles CO2 1 mole of C2H2 = 26.0 g of C2H2 1 mole C2H2 and 26.0 g C2H2 26.0 g CO2 1 mole C2H2 STEP 4 Setup to calculate g of O2: 75.0 g CO2 x 1 mole CO2 x 2 moles C2H2 x 26.0 g C2H2 44.0 g CO2 4 moles CO2 1 mole C2H2 = 22.2 g C2H2 (3)

Learning Check How many grams H2O are produced when 35.8 g of C3H8 react by the following equation? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 1. 14.6 g of H2O 2. 58.4 g of H2O 3. 117 g of H2O

Solution 2) 58.4 g H2O STEP 1 Given: 38.5 g of C3H8 Need: mass (g) of H2O STEP 2 Write a plan to calculate grams of H2 O: g of C3H8 moles of moles of g of H2O C3H8 H2O STEP 3 Write conversion factors: 1 mole of C3H8 = 4 moles of H2O 4 moles H2O and 1 mole C3H8 1 mole C3H8 4 moles H2O

Solution (continued) STEP 3 Write conversion factors (continued): 1 mole of C3H8 = 44.1 g of C3H8 1 mole C3H8 and 44.1 g C3H8 44.1 g C3H8 1 mole C3H8 1 mole of C2H2 = 26.0 g of C2H2 1 mole H2O and 18.0 g H2O 18.0 g H2O 1 mole H2O STEP 4 Setup to calculate g of H2O: 35.8 g C3H8 x 1 mole C3H8 x 4 moles H2O x 18.0 g H2O 44.1 g C3H8 1 mole C3H8 1 mole H2O = 58.4 g of H2O (2)