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

i just calculated it all out, and not including weekends, and all the different days off. as of this school year (2018-2019) students are spending 2198 hours in school this year

the awnser is group 12