Determine the mass fraction of alkali obtained as a result of the interaction of 9.9 g of water and 10.34 g

Determine the mass fraction of alkali obtained as a result of the interaction of 9.9 g of water and 10.34 g of potassium oxide.

The reaction for the synthesis of alkali from oxide is described by the following chemical reaction equation:

K2O + H2O = 2KON;

In this reaction, 1 mole of oxide reacts with 1 mole of water. This produces 2 moles of alkali.

Let’s calculate the chemical amount of a substance that is in 10.34 grams of potassium oxide.

M K2O = 39 x 2 + 16 = 94 grams / mol;

N K2O = 10.34 / 94 = 0.11 mol;

This reaction also requires 0.11 mol of water.

The weight of the water will be:

M H2O = 2 + 16 = 18 grams / mol;

N H2O = 0.11 x 18 = 1.98 grams;

We calculate the weight of 0.11 x 2 = 0.22 mol of alkali.

M KOH = 39 + 16 + 1 = 56 grams / mol;

m KOH = 0.22 x 56 = 12.32 grams;

Let’s find the mass fraction of alkali in the solution. The mass of the solution will be 9.9 + 10.34 = 20.24 grams;

With alkali = 12.32 / 20.24 = 0.6087 = 60.87%;