When fusing 150 grams of limestone with silicon oxide 4, 145 grams of CaSiO3

When fusing 150 grams of limestone with silicon oxide 4, 145 grams of CaSiO3 were formed. Determine the mass fraction of CaCo3 in 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 = 145/116 = 1.25 mol;

Determine the weight of 1.25 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 = 1.25 x 100 = 125 grams;

Mass fraction of calcium carbonate in limestone is 125/150 = 0.8333 = 83.33%;