How much oxygen is needed to burn 11.2 liters of butane?

Let’s execute the solution:
1. We compose the equation according to the condition of the problem:
V = 11.2 liters. X l. -?
2С4Н10 + 13О2 = 8СО2 + 10Н2О + Q – butane combustion, carbon dioxide, water, heat are released;
2. Proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (C4H10) – 11.2 liters. hence, X mol (C4H10) = 1 * 11.2 / 22.4 = 0.5 mol;
0.5 mol (C4H10) – X mol (O2);
-2 mol -13 mol from here, X mol (O2) = 0.5 * 13/2 = 3.25 mol.
3. Find the volume of O2:
V (O2) = 3.25 * 22.4 = 72.8 liters.
Answer: 72.8 liters are required to carry out the reaction. butane.