How much silicon oxide (IV) and sodium oxide is required to obtain sodium silicate weighing 36.6 g?

Let’s write the reaction equation:

Na2O + SiO2 (t) = Na2SiO3

Let’s find the amount of sodium silicate substance:

v (Na2SiO3) = m (Na2SiO3) / M (Na2SiO3) = 36.6 / 122 = 0.3 (mol).

According to the reaction equation, 1 mol of Na2SiO3 is formed from 1 mol of Na2O and 1 mol of SiO2, therefore:

v (Na2O) = v (SiO2) = v (Na2SiO3) = 0.3 (mol).

Thus, the required mass of silicon oxide (IV):

m (SiO2) = v (SiO2) * M (SiO2) = 0.3 * 60 = 18 (g),

and the required mass of sodium oxide:

m (Na2O) = v (Na2O) * M (Na2O) = 0.3 * 62 = 18.6 (g).