Determine the mass of Al that will be consumed for the reaction of O2 with a mass of 11.2 liters.

Let’s implement the solution:

Let’s make the equation:
Xr -? V = 11.2 l;

4Al + 3O2 = 2Al2O3 – compounds, aluminum oxide was obtained.

Let’s make the calculations:
M (Al) = 26.9 g / mol;

M (O2) = 32 g / mol.

Proportions:
1 mol of gas at normal level – 22.4 liters;

X mol (O2) – 11.2 liters from here, X mol (O2) = 1 * 11.2 / 22.4 = 0.5 mol;

0.5 mol (O2) – X mol (Al);

-3 mol -4 mol from here, X mol (Al) = 0.5 * 4/3 = 0.6 mol.

Determine the mass of the metal:
m (Al) = Y * M = 0.6 * 26.9 = 16.14 g

Answer: you need aluminum weighing 16.14 g