Find the mass of phosphorus that is formed by the interaction of 120 g of calcium phosphate

Find the mass of phosphorus that is formed by the interaction of 120 g of calcium phosphate containing 20% impurities with sand and coke if the mass fraction of the phosphorus yield is 85%?

Let’s find the mass of calcium phosphate without impurities.

100% – 20% = 80%.

120 g – 100%,

X – 80%,

X = (120 g × 80%): 100% = 96 g.

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 = 0.31 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 2 mol of phosphorus per 1 mol of Ca3 (PO4) 2. The substances are in quantitative ratios of 1: 2.

The amount of phosphorus will be 2 times more than the amount of Ca3 (PO4) 2.

n (P) = 2 n (Ca3 (PO4) 2) = 0.31 × 2 = 0.62 mol.

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

m = n × M,

M (P) = 31 g / mol.

m = 0.6 mol × 31 g / mol = 19.22 g.

19.22 g was obtained according to calculations. According to the condition of the problem, the yield is 85%.

19.22 – 100%,

X – 85%.

X = (19.22 × 85%): 100% = 16.34 g

Answer: 16.34 g.