# Silicon oxide 4 weighing 30 grams was fused with 30 grams of sodium hydroxide

**Silicon oxide 4 weighing 30 grams was fused with 30 grams of sodium hydroxide to determine the mass of sodium silicate that can be obtained.**

Let’s execute the solution:

In accordance with the condition of the problem, we compose the equation of the process:

SiO2 + 2NaOH = Na2SiO3 + H2O – ion exchange, sodium silicate obtained;

Calculations:

M (SiO2) = 60 g / mol;

M (NaOH) = 39.9 g / mol;

M (Na2SiO3) = 121.8 g / mol.

Let’s determine the amount of starting materials:

Y (SiO2) = m / M = 30/60 = 0.5 mol (deficient substance);

Y (Na2SiO3) = 0.5 mol since the amount of substances is 1 mol;

Y (NaOH) = m / M = 30 / 39.9 = 0.75 mol (substance in excess).

Calculations are carried out for the substance in deficiency.

Find the mass of the product:

m (Na2SiO3) = Y * M = 0.5 * 121.8 = 60.9 g

Answer: the mass of sodium silicate is 60.9 g