The mass of sodium hydroxide formed by the interaction of 12.4 g of sodium oxide with water

The mass of sodium hydroxide formed by the interaction of 12.4 g of sodium oxide with water in accordance with the reaction equation Na2O + H2O = 2NaOH is ..

Given:
m (Na2O) = 12.4 g

To find:
m (NaOH) -?

1) Write the reaction equation:
Na2O + H2O => 2NaOH;
2) Calculate the molar mass of Na2O:
M (Na2O) = Mr (Na2O) = Ar (Na) * N (Na) + Ar (O) * N (O) = 23 * 2 + 16 * 1 = 62 g / mol;
3) Calculate the amount of substance Na2O:
n (Na2O) = m (Na2O) / M (Na2O) = 12.4 / 62 = 0.2 mol;
4) Determine the amount of NaOH substance:
n (NaOH) = n (Na2O) * 2 = 0.2 * 2 = 0.4 mol;
5) Calculate the mass of NaOH:
m (NaOH) = n (NaOH) * M (NaOH) = 0.4 * 40 = 16 g.

Answer: The mass of NaOH is 16 g.