# 5 kg of sand containing 10% impurities was fused with an excess of sodium hydroxide to obtain

**5 kg of sand containing 10% impurities was fused with an excess of sodium hydroxide to obtain 9 kg of product. Calculate the sodium silicate yield as a percentage of the theoretically possible**

Let’s write the reaction equation:

SiO2 + 2NaOH → Na2SiO3 + H2O (conditions: temperature; sodium silicate).

Let’s find the mass of silicon oxide (sand) without impurities:

m (SiO2) = (mtot. * w (SiO2)) / 100% = (5000 g * 90%) / 100% = 4500 g.

Let’s find the amount of silicon oxide substance:

n (SiO2) = m (SiO2) / M (SiO2) = 4500 g / 60 g / mol = 75 mol.

Let’s find the theoretical mass of sodium silicate:

n (Na2SiO3) = n (SiO2) = 75 mol.

m (Na2SiO3) = n (Na2SiO3) * M (Na2SiO3) = 75 mol * 122 g / mol = 9150 g = 9.15 kg.

Let’s calculate the yield of sodium silicate:

n (Na2SiO3) = (mpract. (Na2SiO3) * 100%) / mtheoret. (Na2SiO3) = (9 kg * 100%) / 9.15 kg = 98.36%.