Burns 5 grams of aluminum. How many liters of air did the reaction take?

1. Combustion of aluminum proceeds according to the equation:

4Al + 3O2 = 2Al2O3;

2.Calculate the chemical amount of burning aluminum:

n (Al) = m (Al): M (Al) = 5: 27 = 0.1852 mol;

3. Let’s make a proportion from the reaction equation:

with 4 mol of Al reacts – 3 mol of O2;

and with 0.1852 – x, where x is the chemical amount of oxygen, we express x:

x = (0.1852 * 3): 4 = 0.1389, that is, n (O2) = 0.1389 mol;

4.find the volume of oxygen:

V (O2) = n (O2) * Vm;

V (O2) = 0.1386 * 22.4 = 3.11 dm3;

5. The air contains 21% oxygen by volume. Let’s find the volume of air required for combustion:

V (air) = V (O2): ϕ (O2);

V (air) = 3.11: 0.21 = 14.8 dm3.

Answer: 14.8 dm3.