What volume of air is required for combustion of technical sulfur with a mass of 150 g,

What volume of air is required for combustion of technical sulfur with a mass of 150 g, containing 4% of impurities? The reaction product is sulfur oxide (4). The volume fraction of oxygen in the air is 21%.

1. Let’s compose the equation of a chemical reaction.

S + O2 = SO2.

2. Let’s calculate the amount of burned pure sulfur.

ω (S) = 100% – ω (impurities) = 100% – 4% = 96%.

m (S) = m (mixture) * ω (S) / 100% = 150 g * 96% / 100% = 144 g.

n (S) = m (S) / M (S) = 144 g / 32 g / mol = 4.5 mol.

3. Using the reaction equation, we find the amount and, as a result, the volume of air required to burn a given amount of sulfur.

n (O2) = n (S) = 4.5 mol.

V (O2) = n (O2) * Vm = 4.5 mol * 22.4 l / mol = 100.8 l.

V (air) = V (O2) * 100% / ω (O2) = 100.8 l * 100% / 21% = 480 l.

Answer: V (air) = 480 liters.