Calculate the mass of phosphorite containing 62% of calcium orthophosphate required to obtain 3.1 kg

Calculate the mass of phosphorite containing 62% of calcium orthophosphate required to obtain 3.1 kg of phosphorus, if the phosphorus yield is 92% of the theoretically possible.

Given:
ω (Ca3 (PO4) 2) = 62%
m pract. (P) = 3.1 kg = 3100 g
η (P) = 92%

To find:
m (phosphorite) -?

1) m theor. (P) = m practical (P) * 100% / η (P) = 3100 * 100% / 92% = 3369.57 g;
2) n theor. (P) = m theore. (P) / M (P) = 3369.57 / 31 = 108.7 mol;
3) n (P in Ca3 (PO4) 2) = n theory. (P) = 108.7 mol;
4) n (Ca3 (PO4) 2) = n (P in Ca3 (PO4) 2) / 2 = 108.7 / 2 = 54.35 mol;
5) m (Ca3 (PO4) 2) = n (Ca3 (PO4) 2) * M (Ca3 (PO4) 2) = 54.35 * 310 = 16848.5 g;
6) m (phosphorite) = m (Ca3 (PO4) 2) * 100% / ω (Ca3 (PO4) 2) = 16848.5 * 100% / 62% = 27175 g = 27.1 kg.

Answer: The mass of phosphorite is 27.1 kg.