Question

The next three questions are based on the combustion of propane gas (C₃H₈) with oxygen to produce carbon dioxide and water vapor.

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Stu Dent measures exactly 5.10 g of C₃H₈ and burns it with excess oxygen gas. What is his theoretical yield of carbon dioxide gas?  What is his theoretical yield of water vapor?  (Both answers should have 3 significant digits and units of g)

Stu measures his experimental products. He finds 13.3 g CO₂ and 6.94 g H₂O formed. What is the percent yield for carbon dioxide?  What is the percent yield for water?  (Both answers should have 3 significant digits with units of %)

Calculate Stu's percent error for his two products. (Both answers should have 3 significant digits with units of %)  

Percent error for carbon dioxide?  Percent error for water?

Answer 1

From the reaction

44 g C3H8 produces = 132 g CO2

5.10 g C3H8 produces = 132*5.10/44 = 15.3 g CO2

theoretical yield of carbon dioxide gas = 15.3 g

From the reaction

44 g C3H8 produces = 18*4 = 72 g H2O

5.10 g C3H8 produces = 72*5.10/44 = 8.345 g H2O

theoretical yield of water vapor = 8.345 g

Percent yield of CO2 = actual yield*100/theoretical yield

= 13.3*100/15.3 = 86.928%

Percent yield of H2O = 6.94*100/8.345 = 83.164%

Percent error for CO2 = (15.3-13.3)*100/15.3 = 13.072%

Percent error for H2O =( 8.345-6.94)*100/8.345 = 16.836%