# When fusing 170 g of soda containing 92% sodium carbonate with sand, 175 g

**When fusing 170 g of soda containing 92% sodium carbonate with sand, 175 g of sodium silicate was obtained, what is the practical product yield (in%) of theoretically possible**

Find the mass of sodium carbonate in 170 g of soda.

170% – 100%,

x% -92%.

x = (170% × 92%): 100% = 156.4 g.

Find the amount of sodium carbonate substance by the formula:

n = m: M.

M (Na2CO3) = 23 × 2 + 12 + 16 × 3 = 106 g / mol.

n = 156.4 g: 106 g / mol = 1.48 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

Na2CO3 + SiO2 = Na2 SiO3 + CO2.

According to the reaction equation, there is 1 mole of sodium silicate per 1 mole of soda. The substances are in quantitative ratios of 1: 1.

The amount of substance will be the same.

n (Na2CO3) = n (Na2 SiO3) = 1.48 mol.

Find the mass of sodium silicate by the formula:

m = n × M,

M (Na2 SiO3) = 23 × 2 + 28 + 16 × 3 = 122 g / mol.

m = 1.48 mol × 122 g / mol = 180.56 g.

180.56 g was calculated (theoretical yield).

According to the condition of the problem, sodium silicate obtained 175 g.

Let’s find a practical way out of the product.

180.56 g – 100%,

175 g – x%,

x = (175 × 100%): 180.56 g = 96.92%.

Answer: 96.92%.