Determine the amount of aluminum substance that was mixed with water if 22.4 liters of hydrogen were released.

Let’s execute the solution:
1. 2Al + 6H2O = 2Al (OH) 3 + 3H2 – OBP, hydrogen is released;
2. We make up the proportions:
1 mol of gas at normal level – 22.4 liters;
X mol (H2) – 22.4 liters. hence, X mol (H2) = 1 mol;
X mol (Al) – 1 mol (H2);
– 2 mol – 3 mol from here, X mol (Al) = 1 * 2/3 = 0.66 mol.
Answer: for ORP, aluminum is required in the amount of 0.66 mol.