How many liters of oxygen will it take to burn 6.4 grams of methanol?

Let’s write the reaction equation:

2CH3OH + 3O2 (t) = 2CO2 + 4H2O

Let’s find the amount of methanol substance:

v (CH3OH) = m (CH3OH) / M (CH3OH) = 6.4 / 32 = 0.2 (mol).

According to the reaction equation, 2 mol of CH3OH reacts with 3 mol of O2, therefore:

v (O2) = v (CH3OH) * 3/2 = 0.2 * 3/2 = 0.3 (mol).

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

V (O2) = v (O2) * Vm = 0.3 * 22.4 = 6.72 (l).

Answer: 6.72 l.