A mixture of 5.6 liters of hydrogen and 11.2 liters of oxygen was ignited. What gas and in what volume

A mixture of 5.6 liters of hydrogen and 11.2 liters of oxygen was ignited. What gas and in what volume was left after the reaction?

1.2H2 + O2 = 2H2O;

2.n (H2) = V (H2): Vm;

n (H2) = 5.6: 22.4 = 0.25 mol;

3.n (O2) = V (O2): Vm;

n (O2) = 11.2: 22.4 = 0.5 mol;

4.Calculate the amount of water by hydrogen:

n (H2O) = n (H2) = 0.25 mol;

5.Calculate the amount of water by oxygen:

n (H2O) = n (O2) * 2 = 0.5 * 2 = 1 mol;

since a smaller value of the chemical amount of water is obtained when calculating through hydrogen, then hydrogen is in short supply and all calculations must be carried out using it.

6. Find the amount of reacted oxygen:

nproreag (О2) = n (H2): 2 = 0.25: 2 = 0.125 mol;

7.Calculate the amount of remaining oxygen:

nres (О2) = n (О2) – nproreag (О2) = 0.25 – 0.125 = 0.125 mol;

8.Let’s find the volume of unreacted О2:

Vost (О2) = nres (О2) * Vm = 0.125 * 22.4 = 2.8 liters.

Answer: O2 with a volume of 2.8 liters.