In the following reaction 8 grams of ethane are burned and 11 grams of CO2 are collected. What is the percent yield? 2C2H6+7O2 = 4CO2 + 6H2O

47% yield.

First, let's determine how many moles of ethane was used and how many moles of CO2 produced. Start with the respective atomic weights.

Atomic weight carbon = 12.0107

Atomic weight hydrogen = 1.00794

Atomic weight oxygen = 15.999

Molar mass C2H6 = 2 * 12.0107 + 6 * 1.00794 = 30.06904 g/mol

Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol

Moles C2H6 = 8 g / 30.06904 g/mol = 0.266054387 mol

Moles CO2 = 11 g / 44.0087 g/mol = 0.249950578 mol

Looking at the balanced equation, for every 2 moles of C2H6 consumed, 4 moles of CO2 should be produced. So at 100% yield, we should have 0.266054387 / 2 * 4 = 0.532108774 moles of CO2. But we only have 0.249950578 moles, or 0.249950578 / 0.532108774 = 0.46973587 =

46.973587% of what was expected.

Rounding to 2 significant figures gives 47% yield.