Butane gas is used as the fuel in disposable lighters. It burn in oxygen to form carbon dioxide gas and water vapor. What mass of carbon dioxide is produced when 53.3 g of butane is burned in 103.9 g of oxygen

mass of carbon dioxide produced = 87.9153846154 g ≈ 87.92 g

Explanation:

The chemical equation is as follow

Butane = C4H10

oxygen = O2

carbon dioxide = CO2

water = H2O

C4H10 + O2 → CO2 + H2O

The equation should be balanced before solving it.

2C4H10 + 13O2 → 8CO2 + 10H2O

molar mass of 13 mole of oxygen = 416 g

molar mass of 8 mole of carbon dioxide = 256 + 96 = 352 g

416 g of oxygen was needed to produce 352 g of carbon dioxide

103.9 g will produce?

cross multiply

mass of carbon dioxide produced = 103.9 * 352/416

mass of carbon dioxide produced = 103.9 * 352/416

mass of carbon dioxide produced = 36572.8 / 416

mass of carbon dioxide produced = 87.9153846154 g ≈ 87.92 g