What mass of phosphorus can be obtained from 31 g of natural calcium phosphate containing 10% impurities?

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

100% – 10% = 90%.

2. Let’s find what mass of calcium phosphate is 90%.

31 g – 100%,

x – 90%,

x = (90% × 31g): 100%,

x = 27.9 g.

3.Let’s find the amount of calcium phosphate substance.

M (Ca3 (PO4) 2) = 40 × 3 + 2 (31 + 16 × 4) = 120 + 2 × 95 = 310 g / mol

n (Ca3 (PO4) 2) = 27.9: 310 g / mol = 0.09 mol.

We compose the reaction equation for obtaining phosphorus from calcium phosphate:

2 Сa3 (PO4) 2 + 6 SiO2 + 10C = CaSiO3 + 4P + 10CO.

For 2 moles of calcium phosphate, there are 4 moles of phosphorus, that is, they are in quantitative ratios of 2: 4, then the amount of phosphorus substance will be 2 times more than the amount of calcium phosphate substance.

n (P) = 2 n (Ca3 (PO4) 2) = 0.09 × 2 = 0.18 mol.

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

m = Mn,

M (P) = 31 g / mol.

m = 31 g / mol × 0.18 = 5.58 g.

Answer: m = 5.58 g.