Question

The balanced equation for the combustion of butane is: If the combustion of 53.1 g of C4H10 produces 88.7 g of CO2. What is the percent yield of the reaction? (Assume oxygen is in excess.) %

Answer 1

The balanced equation for the combustion of butane is :

2C4H10 + 13O2 -----> 8CO2 + 10H2O

Molar mass of butane = 58.12 g/mol

53.1 g of butane is 53.1/58.12 = 0.913 mol

2 moles of butane give 8 moles of CO2,

0.913 mol of butane should give 0.913*8/2 = 3.654 mol of CO2

Molar mass of CO2 = 44 g/mol

Mass of 3.654 mol of CO2 = 44*3.654 = 160.79 g

Yield = 160.79- 88.7/160.79 *100 = 44.8%