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

The answer is celsius