If a particular ore contains 58.9 % calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

1 mole of phosphorous contains 31 g / mol

Thus, 1000 g will contain 1000/31 moles

The RFM of calcium phosphate (Ca₃ (PO₄) ₂ = 310 G

1 Mole of calcium phosphate contains 2 moles of P

Therefore, the number of moles of Calcium phosphate that will contain 1000/31 moles will be;

(1/2) * (1000/31) = 1000/62 moles

but 1 mole of calcium phosphate = 310 g

thus, (1000/62) moles calcium phosphate will contain;

= (1000/62) * 310

= 5000 g or 5 kg

Therefore, for 5 kg of calcium phosphate to be processed we will need

(5000 / 58.9) * 100

= 8488.96 g or 8.489 kg of the ore

Thus, a minimum of 8.489 kg of the ore must be processed to yield 1 kg of phosphorus