Find the volume of oxygen and its amount required for combustion a) 5.8 l (heptane) b) 1.6 g (methane)

a) C7H16 + 11O2 = 7CO2 + 8H2O
Amount of heptane: n = V / Vm, where Vm = 22.4 l / mol – molar volume. n (C7H16) = 5.8 L / 22.4 L / mol = 0.26 mol.
According to the reaction, 11 mol of O2 and 1 mol of heptane, n (O2) = 0.26 * 11 = 2.86 mol.
Oxygen volume: V = n * Vm = 2.86 mol * 22.4 l / mol = 64.064 l.
b) CH4 + 2O2 = CO2 + 2H2O
The amount of methane: n = m / M, where M = 16 g / mol (according to the table) n (CH4) = 1.6 g / 16 g / mol = 0.1 mol.
According to the reaction of 1 mol of CH4 with 2 mol of O2, n (O2) = 0.1 * 2 = 0.2 mol.
V (O2) = 0.2 mol * 22.4 L / mol = 4.48 L.