Phosphorus is obtained primarily from ores containing calcium phosphate.

If a particular ore contains 53.1% calcium phosphate, what minimum mass of the ore must be processed to obtain 3.57 kg of phosphorus?

33.7 kg

Explanation:

Let's consider calcium phosphate Ca₃ (PO₄) ₂.

The molar mass of Ca₃ (PO₄) ₂ is 310.18 g/mol and the molar mass of P is 30.97 g/mol. In 1 mole of Ca₃ (PO₄) ₂ (310.18 g) there are 2 * 30.97 g = 61.94 g of P. The mass of Ca₃ (PO₄) ₂ that contains 3.57 kg (3.57 * 10³ g) of P is:

3.57 * 10³ g * (310.18 g Ca₃ (PO₄) ₂/61.94 g P) = 1.79 * 10⁴ g Ca₃ (PO₄) ₂

A particular ore contains 53.1% calcium phosphate. The mass of the ore that contains 1.79 * 10⁴ g of Ca₃ (PO₄) ₂ is:

1.79 * 10⁴ g Ca₃ (PO₄) ₂ * (100 g Ore / 53.1 g Ca₃ (PO₄) ₂) = 3.37 * 10⁴ g Ore = 33.7 kg Ore