Mixed 64 g of oxygen and 22.4 liters of sulfur oxide (4). Determine the mass and volume fraction

Mixed 64 g of oxygen and 22.4 liters of sulfur oxide (4). Determine the mass and volume fraction of the components in the mixture.

Given:
m (O2) = 64 g
V (SO2) = 22.4 l

Find:
ω (O2) -?
ω (SO2) -?
φ (O2) -?
φ (SO2) -?

1) n (SO2) = V (SSO22) / Vm = 22.4 / 22.4 = 1 mol;
2) M (SO2) = Mr (SO2) = Ar (S) * N (S) + Ar (O) * N (O) = 32 * 1 + 16 * 2 = 64 g / mol;
3) m (SO2) = n (SO2) * M (SO2) = 1 * 64 = 64 g;
4) m (mixture) = m (O2) + m (SO2) = 64 + 64 = 128 g;
5) ω (O2) = m (O2) * 100% / m (mixture) = 64 * 100% / 128 = 50%;
6) ω (SO2) = 100% – ω (O2) = 100% – 50% = 50%;
7) M (O2) = Mr (O2) = Ar (O) * N (O) = 16 * 2 = 32 g / mol;
8) n (O2) = m (O2) / M (O2) = 64/32 = 2 mol;
9) V (O2) = n (O2) * Vm = 2 * 22.4 = 44.8 L;
10) V (mixture) = V (O2) + V (SO2) = 44.8 + 22.4 = 67.2 l;
11) φ (O2) = V (O2) * 100% / V (mixture) = 44.8 * 100% / 67.2 = 66.7%;
12) φ (SO2) = 100% – φ (O2) = 100% – 66.7% = 33.3%.

Answer: Mass fraction of O2 is 50%; SO2 – 50%; volume fraction of O2 – 66.7%; SO2 – 33.3%.




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