2 C4H10 + 13 O2 → → 8 CO2 + 10 H2O How many grams of CO2 are produced if 29.3 grams of C4H10 reacts with 76.7 grams of O2?

Answer 1

First, figure out how many moles of each reactant and determine the limiting reactant:

29.3 g / 58.12 g/mol = 0.504 mol of C4H10 (butane)

76.7 g / 16.00 g/mol = 4.79 mol of O2

According to the balanced chemical equation, you need 13 moles of O2 to react with 2 moles of butane.

4.79 mol O2(2 mol butane / 13 mol O2) = 0.7375 mol of butane equivalent

This means butane is your limiting reactant, so all 0.504 mol will be reacted and there will be excess oxygen.

Now determine how much CO2 you get by the mol-to-mol ratio of butane to CO2:

According to the chemical equation, when 2 moles of butane react, you get 8 moles of CO2.

0.504 mol butane(8 mol CO2 / 2 mol butane) = 2.02 mol of CO2