Help with a question regarding Gravimetric Calculations

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The discussion focuses on gravimetric calculations related to the industrial preparation of elemental phosphorus from phosphate rock. The participant successfully balanced the chemical equations for the reactions involving calcium phosphate and silicon dioxide, yielding P4O10 and elemental phosphorus. They calculated that 69.2% of the original phosphate rock sample consisted of Ca3(PO4)2 based on the yield of 42.5 kg of P4 from 307 kg of rock. The calculations were verified through mole conversions and mass relationships, confirming the accuracy of the approach taken.

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This is the question I have been working through and I just want to know if I am on the right track. Any advice is appreciated.

10. The major industrial method for the preparation of elemental phosphorus is heating a mixture of phosphate rock, carbon, and sand in an electric furnace. The chemical reactions can be represented by

Ca3(PO4)2 + SiO2 → CaSiO3 + P4O10

and

P4O10 + C → P4 + CO

a. First balance the equations.

b. Then calculate the percentage of Ca3(PO4)2 initially present in the rock sample if 307 kg of phosphate rock yields 42.5 kg of elemental P4.

A) I balanced the equations:

2Ca3(PO4)2 + 6SiO2 → 6CaSiO3 + P4O10

and

P4O10 + 10C → P4 + 10CO

B) Then I started to work backwards to figure out part B, below are the steps I took.

moles of P4 = 42.5kg x 1mol P4/ 123.88 g = 343.07 mol

Moles P4O10 = 343.07 mol x 1mol P4O10/1 mol P4 = 343.07 mol

kg P4O10 = 343.07 mol x 283.88g/ 1 mol P4O10 = 97.4 kg P4O10

moles Ca3(PO4)2 = 343.07 mol P4O10 x 2mol Ca3(PO4)2/1mol P4O10 = 686.14 mol

g Ca3(PO4)2 = 686.14 mol Ca(PO4)2 x 309.94 g/1mol Ca(PO4)2 = 212.66 kg Ca3(PO4)2

Percent of Ca3(PO4)2 present in original sample = 212.66 kg/307 kg x 100% = 69.2%
 
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Looks OK to me. Note: to simplify calculations whole process can be written as

Ca3(PO4)2 → 2P

(look at the number of atoms of P). While it is not balanced it correctly preserves the ratio between the starting substance and the product you are interested in, making calculations much faster.
 

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