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.
