93.4k views
1 vote
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

User Ghost Ops
by
7.3k points

1 Answer

1 vote

Answer:

285 g of P₄

Step-by-step 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).(75gCa_(3)(PO_(4))_(2))/(100g(phosphorite)) .(1molCa_(3)(PO_(4))_(2))/(310gCa_(3)(PO_(4))_(2)) .(1molP_(4))/(2molCa_(3)(PO_(4))_(2)) .(124gP_(4))/(1molP_(4)) =285gP_(4)

User Euphemia
by
6.4k points