How much air is needed to burn 40 grams of barium containing 5% impurities?

| June 11, 2021 | education

Let’s write down the reaction scheme and arrange the coefficients:
2Ва + О2 = 2ВаО
Let us find the mass of barium without impurities: m (Ba) = m (in-va) * (1 – w), where m (in-va) is the mass of the substance with impurities, w is the mass fraction of impurities
m (Ba) = 40 * (1 – 0.05) = 38 g
Determine the amount of barium substance n (Ba) = m (Ba) / M (Ba), where M (Ba) = 137 g / mol
n (Ba) = 38/137 = 0.277 mol
According to the reaction equation for 2 mol of barium, 1 mol of oxygen /
Thus, the amount of oxygen substance n (O2) = n (Ba) / 2 = 0.277 / 2 = 0.1385 mol
Oxygen volume V (O2) = n (O2) * Vm, where Vm = 22.4 l / mol – molar volume
V (O2) = 0.1385 * 22.4 = 3.1024 L
Taking into account the volumetric oxygen content in the air (according to reference data) φ = 21%, we find the volume of air:
Vair = V (O2) / φ = 3.1024 / 0.21 = 14.77 l



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