An excess of sodium silicate solution was added to a solution containing 14.6 g of hydrochloric acid

An excess of sodium silicate solution was added to a solution containing 14.6 g of hydrochloric acid. What amount of silicic acid substance was formed as a result of the reaction?

During the reaction of sodium silicate and hydrochloric acid, the following chemical reaction occurs:

Na2SiO3 + 2HCl = 2NaCl + H2SiO3;

Let’s calculate the chemical amount of hydrochloric acid:

For this purpose, we divide its weight by the mass of 1 mole of the substance.

M HCl = 1 + 35.5 = 36.5 grams / mol;

n HCl = 14.6 / 36.5 = 0.4 mol;

With 0.4 mol of hydrochloric acid, 0.4 / 2 = 0.2 mol of sodium silicate will react and an equivalent amount of silicic acid will be obtained.

Let’s define its mass.

To do this, multiply the amount of the substance by its molar weight.

M H2SiO3 = 2 + 28 + 16 x 3 = 78 grams / mol;

m H2SiO3 = 0.2 x 78 = 15.6 grams;