When fusing natural limestone with a mass of 150 g with silicon oxide (4)

When fusing natural limestone with a mass of 150 g with silicon oxide (4), calcium silicate with a mass of 145 g was formed. Determine the mass fraction of carbonate in natural 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 synthesized.

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

M CaSiO3 = 40 + 28 + 16 x 3 = 116 grams / mol; N CaSiO3 = 145/116 = 1.25 mol;

Let’s calculate the mass of 1.25 mol of calcium carbonate.

To do this, 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%;