Help with a question regarding Gravimetric Calculations

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The discussion focuses on the gravimetric calculations related to the preparation of elemental phosphorus from phosphate rock. The user successfully balanced the chemical equations involved in the process and calculated the moles of elemental phosphorus produced. They then worked backwards to determine the mass of calcium phosphate present in the original rock sample, concluding that 69.2% of Ca3(PO4)2 was present. The user notes that while the simplified equation is not balanced, it effectively maintains the ratio needed for quicker calculations. Overall, the calculations appear to be correct and well-structured.
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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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