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What is the maximum amount of P4 that can be produced from 1.0kg of phosphorite if the phorphorite sample is 75 Ca3PO42 by mass?
Wiki User
∙ 12y ago
2Ca3(PO4)2 + 6SiO2 + 10 C -> 6CaSiO3 + P4 + 10 CO
There is 1 kg of solution. 75% of the solution is Ca3(PO4)2.
Change 1kg to grams which is 1000g. 75% of 1000 is 750 Therefore there is 750g of the Ca3(PO4)2.
Find the mole of Ca3(PO4)2 which is the mass divided by the molar mass.
750g/310.4gmol-1 which equals n =2.4mol ...1/2 is the ratio so mol of P4 is 1.2 mol.
to find the mass of P4 you do M * n = 1.2mol * 148.65
therefore the max amount of P4 that can be produced is 150g
Wiki User
∙ 12y ago
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