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 The mole

# How can I use the amount of fuel to predict amount of product in a combustion reaction?

I am having some difficulty solving combustion reaction problems. I don't have problems balancing the equations, but when they are asking something like, 29.1 grams of butane reacts with excess oxygen to form carbon dioxide and water. How many grams of carbon dioxide form? Can you give me some pointers on how to approach this...?
Norm Rackison

Norm, try the following strategy.

1. Write a balanced chemical equation for the combustion reaction. For this problem, it's

C4H10 + (13/2) O2 5 H2O + 4 CO2

or

2 C4H10 + 13 O2 10 H2O + 8 CO2

if fractional coefficients bother you.
2. Find moles of fuel from given information. You know you have 29.1 grams of butane. To convert grams to moles, you must use the molecular weight. For butane, it's 4×12.011 + 10×18 = 58.124, so 58.124 g of butane is equivalent to a mole of butane.
 29.1 g butane ( 1 mol butane58.124 g butane ) = 0.50065 mol butane
(Notice this is an intermediate result so I'm not rounding off to the correct number of significant digits yet.)
3. Convert moles of fuel to moles of product using mole ratios from the balanced chemical equation. From the balanced equation, burning 2 moles of butane produces 8 moles of CO2. So
 0.50065 mol butane ( 8 mol CO22 mol butane ) = 226 mol CO2
4. Convert moles of product to desired units. You must convert moles of CO2 to grams of CO2. The molecular weight of CO2 is required:
 226 mol CO2 ( 44.010 g CO21 mol CO2 ) = 88.1 g CO2
5. Check the answer. It makes sense that the mass of CO2 should be about triple the mass of butane, since 4 CO2 molecules weigh about 4×44 = 176 amu or about 3 times as much as one butane molecule (about 58.1 amu).

You can also check the answer by converting the 88.1 grams of CO2 obtained back into grams of butane. Do you get 29.1 g, within 3 significant figures?

Author: Fred Senese