Question

phosporus can be prepared from calcium phosphate by the following reaction: 2Ca3(PO4)2+6SiO2+10C → 6CaSiO3+P4+10CO Phosphorite is a mineral that contains Ca3(PO4)2 by mass? Assume an excess of the other reactants plus other non-phosphorus-containing compounds. What is the maximum amount of P4 that can be produced from 1.9kg of phosphorite sample is 75% Ca3(PO4)2 by mass? Assume an excess of the other reactants

Answer 1

Answer:

285 g of P₄

Explanation:

Let's consider the following balanced equation.

2 Ca₃(PO₄)₂ + 6 SiO₂ + 10 C → 6 CaSiO₃ + P₄ + 10 CO

We know the following relations:

• 100 g of phosphorite contain 75 g of Ca₃(PO₄)₂
• 2 moles of Ca₃(PO₄)₂ produce 1 mole of P₄
• The molar mass of Ca₃(PO₄)₂ is 310 g/mol

• The molar mass of P₄ is 124 g/mol

Then, for 1.9 kg of phosphorite:

$1900g(phosphorite).\frac{75gCa_{3}(PO_{4})_{2}}{100g(phosphorite)} .\frac{1molCa_{3}(PO_{4})_{2}}{310gCa_{3}(PO_{4})_{2}} .\frac{1molP_{4}}{2molCa_{3}(PO_{4})_{2}} .\frac{124gP_{4}}{1molP_{4}} =285gP_{4}$