By fusing 200 g of limestone with sand, 150 g of calcium silicate were obtained. what percentage

By fusing 200 g of limestone with sand, 150 g of calcium silicate were obtained. what percentage of impurities is in the limestone.

The reaction of interaction of limestone with silicon oxide is described by the following equation:

CaCO3 + SiO2 = CaSiO3 + CO2 ↑;

When 1 mol of limestone and 1 mol of silicon dioxide interact, 1 mol of calcium silicate is obtained.

Let’s find the molar amount of the formed calcium silicate. For this purpose, we divide its weight by its molar weight.

M CaSiO3 = 40 + 28 + 16 x 3 = 116 grams / mol; N CaSiO3 = 150/116 = 1.293 mol;

Determine the weight of 1.293 mol of calcium carbonate.

To do this, we multiply the amount of the substance by its molar weight.

M CaCO3 = 40 + 12 + 16 x 3 = 100 grams / mol;

m CaCO3 = 1.293 x 100 = 129.3 grams;

The mass of impurities is 200 – 129.3 = 70.7 grams;

The mass fraction of impurities is 70.7 / 200 = 0.3535 = 35.35%;