Barium hydroxide was formed from the interaction of barium with water in an amount of 0.2 mol.

Barium hydroxide was formed from the interaction of barium with water in an amount of 0.2 mol. calculate the amount of substances and masses of barium and water that have reacted. what volume of gas was generated?

Given:
n (Ba (OH) 2) = 0.2 mol

To find:
n (Ba) -?
m (Ba) -?
n (H2O) -?
m (H2O) -?
V (gas) -?

1) Ba + 2H2O => Ba (OH) 2 + H2 ↑;
2) n (Ba) = n (Ba (OH) 2) = 0.2 mol;
3) M (Ba) = Ar (Ba) = 137 g / mol;
4) m (Ba) = n (Ba) * M (Ba) = 0.2 * 137 = 27.4 g;
5) n (H2O) = n (Ba (OH) 2) * 2 = 0.2 * 2 = 0.4 mol;
6) M (H2O) = Mr (H2O) = Ar (H) * N (H) + Ar (O) * N (O) = 1 * 2 + 16 * 1 = 18 g / mol;
7) m (H2O) = n (H2O) * M (H2O) = 0.4 * 18 = 7.2 g;
8) n (H2) = n (Ba (OH) 2) = 0.2 mol;
9) V (H2) = n (H2) * Vm = 0.2 * 22.4 = 4.48 liters.

Answer: The amount of substance Ba is 0.2 mol; H2O – 0.4 mol; weight of Ba – 27.4 g; H2O 7.2 g; volume H2 – 4.48 l.