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

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 butane 58.124 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 CO2 2 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 CO2 1 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