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.

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