Calculate the mass of phosphorus that can be obtained by reducing fasforite with a mass of 193.75 g with coal.

Calculate the mass of phosphorus that can be obtained by reducing fasforite with a mass of 193.75 g with coal. The product yield from the theoretically possible should be taken equal to 90%

Let’s find the amount of phosphorite substance by the formula:

n = m: M.

M (Ca3 (PO4) 2) = 40 × 3 + 2 (31 + 16 × 4) = 310 g / mol.

n = 193.75 g: 310 g / mol = 0.625 mol.

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

Ca3 (PO4) 2 + 5 C + 3SiO2 = 3CaSiO3 + 2P + 5CO.

According to the reaction equation, there is 1 mole of phosphorus per mole of phosphorite. Substances are in quantitative ratios 1: 2.

The amount of phosphorus substance will be 2 times more than the amount of phosphorite substance.

n (P) = 2 n (Ca3 (PO4) 2) = 0.625 × 2 = 1.25 mol.

Let’s find the mass of phosphorus by the formula:

m = n × M,

M (P) = 31 g / mol.

m = 1.25 mol × 31 g / mol = 38.75 g.

38.75 g were calculated (theoretical yield).

According to the problem statement, the phosphorus yield was 90%.

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

38.75 g – 100%,

x g – 90%,

x = (38.75 × 90%): 100% = 34.875.

Answer: 34.875.