Question

Solve the following:

What volume of oxygen is required to burn completely 90 dm³ of butane under similar conditions of temperature and pressure?

\[\ce{2C4H10 + 13O2 -> 8CO2 + 10H2O}\]

Answer 1

\[\ce{\underset{2 Vols.}{2C4H10_{(g)}} + \underset{13 Vols.}{13O2}-> \underset{8 Vols.}{8CO2} + 10H2O}\]

2 vols. of butane requires O₂ = 13 vols.

90 dm³ of butane will require O₂ = `13/2 xx 90`

= 585 dm³