Answer:
6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
Explanation:
Given
An ore has 51% Calcium phosphate
To find
The minimum mass of ore to be processed to get 1.00 Kg of phosphorous
First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃(PO₄)₃.
Molar mass of Ca₃(PO₄)₃ is 310 g
Molar mass of P is 31 g
1 mole of Ca₃(PO₄)₃ has 3 atoms of phosphorous
i.e 310 g of Ca₃(PO₄)₃ has 3× 31g of P
= 93 g P
93 g P wil be present in 310 g of Ca₃(PO₄)₃
1 Kg or 1000 g of P will be in (1000÷93) ×310
=3333.33 g of Ca₃(PO₄)₃
but the ore has only 55.1% Ca₃(PO₄)₃
i.e
100 g of Ca₃(PO₄)₃ will have 55.1g Ca₃(PO₄)₃
we need 3333.33g of Ca₃(PO₄)₃
100 g of ore will have 55.1g Ca₃(PO₄)₃
3333.33g of Ca₃(PO₄)₃ will be present in
(3333.33÷ 55.1) ×100
= 6049.60 g of the ore
So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.
