Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (l) At 1.00 atm and 23 oC, what is the volume of carbon dioxide formed by the combustion of 3.4 g of butane?

5.638 L is the volume formed

Explanation:

The reaction for the combustion is:

2 C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (l)

First of all we convert the mass of butane to moles (mass / molar mass)

3.4 g / 58 g/mol = 0.058 moles

As we assume, the oxygen in excess ratio between butane and carbon dioxide is 2:8. Let's make a rule of three:

2 moles of butane can produce 8 moles of carbon dioxide

0.058 moles of butane must produce (0.058. 8) / 2 = 0.234 moles of CO₂

Now we apply the Ideal Gases Law to find out the volume formed.

P. V = n. R. T

1atm. V = 0.234 mol. 0.082 L. atm/mol. K. 296K

V = 0.234 mol. 0.082 L. atm/mol. K. 296K / 1atm = 5.68 L