To 300 liters of air was added 60 liters of nitrogen. Calculate the masses of oxygen and nitrogen that can be obtained from this mixture.
V (air) = 300 l
V (N2) = 60 l
n (N2) = V (N2) / Vm
V (N2) = 300 * 78/100 = 234l + 60 = 294l
V (O2) = 300 * 21/100 = 62l
m (N2) = 28 * 294 / 22.4 = 367.5g
m (O2) = 32 * 62 / 22.4 = 88.6g
Answer: 367.5g, 88.6g
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