Air contains 21% oxygen and 78% nitrogen by volume. How many grams of oxygen will be contained in 300 liters of air?
| May 15, 2021 | education
Given:
φ (O2) = 21%
φ (N2) = 78%
V (air) = 300 l
To find:
m (O2) -?
Decision:
1) V (O2) = φ (O2) * V (air) / 100% = 21% * 300/100% = 63 l;
2) n (O2) = V / Vm = 63 / 22.4 = 2.81 mol;
3) m (O2) = n * M = 2.81 * 32 = 89.9 g.
Answer: The mass of O2 is 89.9 g.
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