How much oxygen is spent on the combustion of 15.6 g of benzene?

Let’s write the reaction equation:

2C6H6 + 15O2 (t) = 12CO2 + 6H2O

Let’s find the amount of benzene substance:

v (C6H6) = m (C6H6) / M (C6H6) = 15.6 / 78 = 0.2 (mol).

According to the reaction equation, 2 mol of C6H6 reacts with 15 mol of O2, therefore:

v (O2) = v (C6H6) * 15/2 = 0.2 * 15/2 = 1.5 (mol).

Thus, the required volume of oxygen, measured under normal conditions (n.o.):

V (O2) = v (O2) * Vm = 1.5 * 22.4 = 33.6 (l).

Answer: 33.6 l.