The gas obtained from the decomposition of sodium nitrate weighing 425 g reacted with another gas

The gas obtained from the decomposition of sodium nitrate weighing 425 g reacted with another gas, which was formed by the action of excess potassium hydroxide on aluminum weighing 45 g. determine the mass of the product obtained.

Let’s write down the reaction equations:
2NaNO3 = 2NaNO2 + O2

2KOH + 2Al + 6H2O = 2K [Al (OH) 4] + 3H2

O2 + 2H2 = 2H2O

Let’s calculate the amount of oxygen:
n (NaNO3) = m (NaNO3) / Mr (NaNO3) = 425/85 = 5 mol;

n (O2) = 0.5n (NaNO3) = 0.5 * 5 = 2.5 mol;

Let’s calculate the amount of hydrogen:
n (Al) = m (Al) / Mr (Al) = 45/27 = 1.67 mol;

n (H2) = 3 / 2n (Al) = 3/2 * 1.67 = 2.5 mol;

Let’s find the mass of the resulting water:
The reaction requires twice as much hydrogen as oxygen. This means that oxygen is in excess, n (H2O) = n (H2) = 2.5 mol;

m (H2O) = n (H2O) * Mr (H2O) = 2.5 * 18 = 45 g.

Answer: 45 g.