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2Ca (PO.)2 + 6SiO2 + 10C - 6Casio, + PA+10CO
Starting with 700.0 grams of calcium phosphate and assuming excess silicon dioxide and carbon,
how many moles of phosphorus will be produced?

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CintiaSalazar

Taking into account the reaction stoichiometry, 0.1129 moles of P₄ is required to react with with 700.0 grams of calcium phosphate.

Reaction stoichiometry

In first place, the balanced reaction is:

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

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Ca₃(PO₄)₂: 2 moles
  • SiO₂: 6 moles
  • C: 10 moles
  • CaSiO₃: 6 moles
  • CO: 10 moles
  • P₄: 1 mole

The molar mass of the compounds is:

  • Ca₃(PO₄)₂: 310 g/mole
  • SiO₂: 60 g/mole
  • C: 12 g/mole
  • CaSiO₃: 116 g/mole
  • CO: 28 g/mole
  • P₄: 124 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Ca₃(PO₄)₂: 2 moles ×310 g/mole= 620 grams
  • SiO₂: 6 moles ×60 g/mole= 360 grams
  • C: 10 moles ×12 g/mole=  120 grams
  • CaSiO₃: 6 moles ×116 g/mole= 696 grams
  • CO: 10 moles ×28 g/mole= 280 grams
  • P₄: 1 mole ×124 g/mole= 124 grams

Moles of phosphorus P₄ produced

The following rule of three can be applied: If by reaction stoichiometry 620 grams of Ca₃(PO₄)₂ produces 1 mole of P₄, 70 grams of Ca₃(PO₄)₂ produces how many moles of P₄?

[tex]moles of P_{4} =\frac{70 grams of Ca_{3}(PO_{4} )_{2}x1 mole of P_{4} }{620 grams of Ca_{3}(PO_{4} )_{2}}[/tex]

moles of P₄= 0.1129 moles

Finally, 0.1129 moles of P₄ is required to react with with 700.0 grams of calcium phosphate.

Learn more about the reaction stoichiometry:

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