Calculate the volume of air required for complete combustion of 6.4 g methanol.

Let’s find the amount of methanol substance (CH3OH).

n = m: M.

M (CH3OH) = 32 g / mol.

n = 6.4 g: 32 g / mol = 0.2 mol.

Let’s compose the reaction equation, find the quantitative relationships of substances.

2СН4О + 3O2 = 2СО2 ↑ + 4H2O.

For 2 mol of methanol there are 4 mol of oxygen. The substances are in quantitative ratios of 2: 3. The amount of oxygen substance will be 1.5 times greater than the amount of methanol substance.

n (О2) = 2n (СН4О) = 0.2 × 1.5 = 0.3 mol.

Let’s find the volume of oxygen.

V = Vn n, where Vn is the molar volume of gas equal to 22.4 l / mol, and n is the amount of substance.

V = 22.4 L / mol × 0.3 mol = 6.72 L.

6.72 L – 21%,

xl – 100%.

X = (6.72 × 100): 21% = 32 liters.

Answer: V = 32 liters.