Question

The combustion of 296 g of propane (C₃H₈) produces 713 g of carbon dioxide. [C₃H₈ + 5O₂ --> 3CO₂ + 4H₂O; C = 12.01 g/mol, H = 1.01 g/mol, O = 16.0 g/mol]

What is the percent yield?


80.5%


124%


0.805%


41.50%

Answer 1

Answer:

80.5%

Explanation:

The percent yield is made up of the theoretical yield and actual yield. The theoretical yield is the amount of product produced from the balanced chemical equation and molar masses. The actual yield is the amount of product produced in an experiment.

To find the percent yield, you need to (1) convert grams C₃H₈ to moles C₃H₈ (via molar mass), then (2) convert moles C₃H₈ to moles CO₂ (via mole-to-mole ratio from equation coefficients), then (3) convert moles CO₂ to grams CO₂ (via molar mass), and then (4) calculate the percent yield. It is important to arrange the conversions/ratios in a way that allows for the cancellation of units.

(Steps 1 - 3)

Molar Mass (C₃H₈): 3(12.01 g/mol) + 8(1.01 g/mol)

Molar Mass (C₃H₈): 44.11 g/mol

Molar Mass (CO₂): 12.01 g/mol + 2(16.0 g/mol)

Molar Mass (CO₂): 44.01 g/mol

1 C₃H₈ + 5 O₂ -----> 3 CO₂ + 4 H₂O

296 g C₃H₈           1 mole            3 moles CO₂           44.01 g
-------------------  x  ----------------  x  ----------------------  x  ----------------  =  886 g CO₂
                               44.11 g            1 mole C₃H₈            1 mole

(Step 4)

                                  Actual Yield
Percent Yield  =  ------------------------------  x  100%
                              Theoretical Yield

                                713 g CO₂
Percent Yield  =  ----------------------  x  100%
                                886 g CO₂

Percent Yield  =  80.5%