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

the seeing the electronic configuration we can say that the element is fluorine.

fluorine is the most electronegative element in the periodic table.

but by only seeing the configuration we cannot say about its reactivity because we don't know if its configuration after excitation or by addition of electrons.

but if not then, it is the lightest halogen and exists as a highly toxic pale yellow diatomic gas at standard conditions. as the most electronegative element, it is extremely reactive: almost all other elements, including some noble gases, form compounds with fluorine.