Question

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

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)

At 1.00 atm and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 1.60 g of butane?

Answer 1

Solution

From the equation it can be concluded that 2 moles of butane form 8 moles of CO₂. So 1 mole of butane form 4 mole of CO₂. Find the amount of moles of butane given:

$$n(C4H10) = m/M = 1.6/58 = 0.0275 (mol)$$

Find the amount of CO₂ formed:

$$n(CO2) = 0.0275*4 = 0.11 (mol)$$

If PV=nRT for each gas at any conditions, find which volume is occupied by 0.11 mol of CO₂ at the conditions indicated in the task:

$$V = nRT/P = nRT$$

$$V = 0.11*(273+23)*0.082 = 2.67 (L)$$

Answer

2.67 L of carbon dioxide are formed by the combustion of 1.60 g of butane.