Q: What is Ca3(PO4)2?

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Ca3(PO4)2 + 3 SiO2 + 5 C = 3 CaSiO3 + 5 CO + 2 P

This appears to be calcium triphosphate, Ca3(PO4)2. In any formula such as this when an element appears without a subscript this means that just one atom of the element is to be understood. Thus we could rewrite this formula as Ca3(P1O4)2. 3 Ca 2 P 8 O -- 13 atoms altogether

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

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It stands for Calcium Phosphate.

Ca3(PO4)2 + 3 SiO2 + 5 C = 3 CaSiO3 + 5 CO + 2 P

This appears to be calcium triphosphate, Ca3(PO4)2. In any formula such as this when an element appears without a subscript this means that just one atom of the element is to be understood. Thus we could rewrite this formula as Ca3(P1O4)2. 3 Ca 2 P 8 O -- 13 atoms altogether

Ca = 40 x 3 = 120 grams P = 31 x 2 = 62 grams O = 16 x 8 = 128 grams Add them together and you have 310 grams. This is the mass in grams of your formula. It is expressed in grams/mole because 310 is the weight of one mole of Ca3(PO4)2.

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