Question

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?

Answer 1

Step 1: Find the % of P in 58.9 g of Calcium phosphate (Ca₃ (PO₄)₂

(molar mass of P)/ (total molar mass of calcium phosphate) x 100
= (61.94 g )/ (310.18 g of calcium phosphate) x 100%

= 19.969% P in calcium phosphate

Step 2: 58.9 % means 58.9 g of the ore is calcium phosphate

So, 19.969% of 58.9 g = 11.7617 g of P in one ore

Step 3: 1 Kg = 1000 g.

We need 1000 grams of P but we get 11.7617 g of P in one ore,

So, number of ores = 1000 g / 11.7617 g = 85.0217 ores

So,  the mass of the total amount of ores =  (85.0217 x 100) g = 8502.17 g = 8502 (round to a whole number)