How much sand is required containing 0.2 mass fraction of impurity to obtain 6.1 kg of sodium silicate.

Given:
ω approx. = 0.2
m (Na2SiO3) = 6.1 kg = 6100 g

To find:
m tech. (SiO2) -?

1) SiO2 + 2NaOH => Na2SiO3 + H2O;
2) M (Na2SiO3) = Mr (Na2SiO3) = Ar (Na) * N (Na) + Ar (Si) * N (Si) + Ar (O) * N (O) = 23 * 2 + 28 * 1 + 16 * 3 = 122 g / mol;
3) n (Na2SiO3) = m / M = 6100/122 = 50 mol;
4) n (SiO2) = n (Na2SiO3) = 50 mol;
5) M (SiO2) = Mr (SiO2) = Ar (Si) * N (Si) + Ar (O) * N (O) = 28 * 1 + 16 * 2 = 60 g / mol;
6) m clean. (SiO2) = n * M = 50 * 60 = 3000 g;
7) ω (SiO2) = 1 – ω approx. = 1 – 0.2 = 0.8;
8) m tech. (SiO2) = m pure. / ω = 3000 / 0.8 = 3750 g.

Answer: The mass of technical SiO2 is 3750 g.