What mass of phosphorus can be obtained from 1 kg of natural calcium phosphate containing 7% impurities?

Given:
m tech. (Ca3 (PO4) 2) = 1 kg = 1000 g
ω (approx.) = 7%

To find:
m (P) -?

1) ω (Ca3 (PO4) 2) = 100% – ω (approx.) = 100% – 7% = 93%;
2) m clean. (Ca3 (PO4) 2) = ω * m tech. / 100% = 93% * 1000/100% = 930 g;
3) M (Ca3 (PO4) 2) = Mr (Ca3 (PO4) 2) = Ar (Ca) * N (Ca) + Ar (P) * N (P) + Ar (O) * N (O) = 40 * 3 + 31 * 2 + 16 * 8 = 310 g / mol;
4) n (Ca3 (PO4) 2) = m pure. / M = 930/310 = 3 mol;
5) n (P) = n (Ca3 (PO4) 2) * 2 = 3 * 2 = 6 mol;
6) M (P) = Ar (P) = 31 g / mol;
7) m (P) = n * M = 6 * 31 = 186 g.

Answer: The mass of P is 186 g.