What volume of air is required for combustion of technical sulfur with a mass of 150 g, containing 4% of impurities?

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. We make a chemical equation, not forgetting to put down the coefficients: 2

S + O2 = SO2.

2. Let’s find the mass fraction of sulfur:

ω (S) = 100% – ω (impurities) = 100% – 4% = 96% (0.96).

3. Let’s find the mass of sulfur:

m (S) = ω * m (mixture) = 0.04 * 150g = 144g.

4. Let’s find the amount of sulfur substance:

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

5. The equation shows:

n (О2) = n (S) = 4.5 mol.

6. Find the volume of oxygen:

V (О2) = Vm * n = 22.4 * 4.5 = 100.8 liters.

7. Let’s find the volume of air:

V = V (О2) / f = 100.8l / 0.21 = 480l.

Answer: V = 480l.