How much ammonia can be oxidized on the catalyst if 56L air is used for this?

univerkov | September 23, 2021 | education

How much ammonia can be oxidized on the catalyst if 56L air is used for this? (the proportion of oxygen in the air is 20%)

4NH3 + 5O2 = 4NO + 6H2O
V (O2) = 56 * 0.2 = 11.2L
n = V / Vm4NH3 + 5O2 = 4NO + 6H2O
V (O2) = 56 * 0.2 = 11.2L
n = V / Vm
n (o2) = 11.2 / 22.4 = 0.5 mol
According to the chemical reaction equation
n (O2) / n (NH3) = 5/4
n (NH3) = 0.4 mol
V = n * Vm
V (NH3) = 0.4 * 22.4 = 8.96L
n (o2) = 11.2 / 22.4 = 0.5 mol
According to the chemical reaction equation
n (O2) / n (NH3) = 5/4
n (NH3) = 0.4 mol
V = n * Vm
V (NH3) = 0.4 * 22.4 = 8.96L

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