How much oxygen and air is required to burn 48 liters of butane?

Note that in the condition of the problem it is indicated under n.u. … This means that under normal conditions, which means the solution is simplified.

first of all, we write down the reaction of combustion of butane.

2 С4Н10 + 13 О2 = 8 СО2 + 10 Н2О

From the reaction it follows that for 1 mol of С4Н10 (butane) 6.5 mol of О2 (oxygen) is required

According to the consequence of Avogadro’s law, 6.5 liters of oxygen are required for 1 liter of butane.

So we get that oxygen is required 48 x 6.5 = 312 (liters) Air contains 21% oxygen. therefore the amount of air will be 312 / 0.21 = 1486 (liters).