When 8.88 g of c6h10 were reacted with excess of o2, 18.9 g of co2 were recovered. given this information, what is the % yield of co2?

66.2% yield  
The balanced equation for the reaction is: 
2C6H10 + 17O2 ==> 12CO2 + 10H2O  
So for every 2 moles of C6H10 consumed, we should get 12 moles of CO2. Or to simplify, for each mole of C6H10, we should get 6 moles of CO2. Now let's calculate the molar mass of C6H10 and CO2 and then determine how many moles of each we really have.
 Atomic weight carbon = 12.0107 
 Atomic weight hydrogen = 1.00794 
 Atomic weight oxygen = 15.999
 Molar mass C6H10 = 6*12.0107 + 10*1.00794 = 82.1436 g/mol
 Molar mass CO2 = 12.0107 + 2*15.999 = 44.0087 g/mol 
 Moles C6H10 = 8.88 g / 82.1436 g/mol = 0.10810337 mol
 Moles CO2 = 18.9 g / 44.0087 g/mol = 0.429460538 mol 
 Since we had 0.10810337 moles of C6H10, we should have gotten
6*0.10810337 = 0.648620221 moles of CO2, but only got 0.429460538 moles. So let's divide the actual yield by the theoretical yield to get the percentage yield. 
0.429460538 / 0.648620221 = 0.662114014 = 66.2%

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