2C8H18+25O2→16CO2+18H2O

Question

1) A person is trying to measure the yield of the gasoline he is buying for his car, so he buys 1kg of gasoline (C8H18, Atomic weight= 114.22 g). In motors gasoline burns out slowly because of the absence of oxygen, so the limiting reagent for this reaction will be ?

2) After 20 minutes, the machine measures 70 moles of oxygen spent, how many moles of CO2 should be produced at this point? *

3) After one hour all the gasoline was burned totally, and the machine measured that only 60 moles of CO2 were produced. How many moles of CO2 should have been produced in theory? *

4) What was the purity ( purity = % yield, in this case) of the gasoline they sold him? *

Answer 1

1.) The limiting reagent is oxygen, as octane is the reagent in excess.

2.) To find the moles of CO2 produced, you must find the mole:mole ratio of CO2 to O2. It is 16/25. So 70x(16/25)= 44.8 moles of CO2

3.) It is given that 1,000g (1 Kg) of Octane was purchased. Divide by the molecular weight to find moles: 1000/114.22=8.755 moles octane. Then you can move to moles of O2 used, or straight to CO2. For O2, multiply by 12.5, which gives you 109.438 moles of O2. Then multiply by (16/25) to get moles CO2, which gives you 70.04 moles CO2.

4.) 60 moles of CO2 are produced, but we know that 70.04 moles should have been theoretically. So to find percent yield, do: (60/70.04)x100%, which equals 85.665% yield.