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

obtain 1.00 kg of phosphorus?

Express your answer with the appropriate units.

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.