Question

When 0.560 mol butane (C4H10) combusts in excess oxygen gas, 75.3 g carbon dioxide (molar mass 44.01 g/mol) forms. What is the percent yield of this reaction? 2 C4H10 (g) + 13 O2 (g) --> 8 CO2 (g) + 10 H2O (g) 98.6% 3.06% 76.4% 1.31%

Answer 1

The percent yield of this reaction is approximately 164.7%.

Step-by-step explanation:

To calculate the percent yield of a reaction, we need to compare the actual yield to the theoretical yield. The theoretical yield is the amount of product that would be obtained if the reaction proceeded perfectly according to stoichiometry. In this case, from the balanced equation, we can see that 2 moles of C4H10 produce 8 moles of CO2. So, the theoretical yield of CO2 can be calculated using the mole ratio:

(75.3 g CO2) x (1 mol CO2/44.01 g CO2) x (2 mol C4H10/8 mol CO2) = 0.340 mol CO2

The percent yield can then be calculated using the formula:

Percent Yield = (Actual yield/Theoretical yield) x 100%

Given that the actual yield is 0.560 mol and the theoretical yield is 0.340 mol, we can substitute these values into the formula:

Percent Yield = (0.560 mol/0.340 mol) x 100% = 164.7%

Therefore, the percent yield of this reaction is approximately 164.7%.