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drTat
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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%
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%