Question

If a particular ore contains 55.1 % calcium phosphate, what minimum mass of the ore must be processed to

Answer 1

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.