Determine the mass of Mg (OH) 2 formed by the interaction of magnesium with water, if 5.6 liters

Determine the mass of Mg (OH) 2 formed by the interaction of magnesium with water, if 5.6 liters of hydrogen were formed.

Given:
V (H2) = 5.6 L

To find:
m (Mg (OH) 2) -?

Solution:
1) Mg + 2H2O => Mg (OH) 2 + H2;
2) n (H2) = V (H2) / Vm = 5.6 / 22.4 = 0.25 mol;
3) n (Mg (OH) 2) = n (H2) = 0.25 mol;
4) m (Mg (OH) 2) = n (Mg (OH) 2) * M (Mg (OH) 2) = 0.25 * 58 = 14.5 g.

Answer: The mass of Mg (OH) 2 is 14.5 g.