Question

Phosphorite is a mineral that contains plus other non-phosphorus-containing compounds. What is the maximum amount of that can be produced from 2.3 kg of phosphorite if the phosphorite sample is 75% by mass? Assume an excess of the other reactants.

Answer 1

Phosphorus can be prepared from calcium phosphate by the following reaction:

$2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)$

Phosphorite is a mineral that contains $Ca_3(PO_4)_2$ plus other non-phosphorus-containing compounds. What is the maximum amount of $P_4$ that can be produced from 2.3 kg of phosphorite if the phorphorite sample is 75% $Ca_3(PO_4)_2$ by mass? Assume an excess of the other reactants.

Answer: Thus the maximum amount of $P_4$ that can be produced is 0.345 kg

Explanation:

Given mass of phosphorite $Ca_3(PO_4)_2$  = 2.3 kg

As given percentage of phosphorite $Ca_3(PO_4)_2$ is = $\frac{75}{100}\times 2.3kg=1.725kg=1725g$

$moles=\frac{\text {given mass}}{\text {Molar mass}}$

${\text {moles of}Ca_3(PO_4)_2=\frac{1725g}{310g/mol}=5.56moles$

$2Ca_3(PO_4)_2(s)+6SiO_2(s)+10C(s)\rightarrow 6CaSiO_3(s)+P_4(s)+ 10CO(g)$

According to stoichiometry:

2 moles of phosphorite gives = 1 mole of $P_4$

Thus 5.56 moles of phosphorite give= $\frac{1}{2}\times 5.56=2.78moles$ of $P_4$

Mass of $P_4=moles\times {\text {Molar mass}}=2.78mol\times 124g/mol=345g=0.345kg$

Thus the maximum amount of $P_4$ that can be produced is 0.345 kg