White phosphorus, P4, is prepared by fusing
calcium phosphate, Ca3(PO4)2, with carbon, C,
and sand, SiO2, in an electric furnace.

2Ca3(PO4)2(s) + 6SiO2(s)
+ 10C(s) --> P4(g)
+ 6CaSiO3(l) + 10CO(g)

How many grams of calcium phosphate are required to give
30.0 g of phosphorus?

Think it's 150. g Ca3(PO4)2 (am I right or wrong? Please explain, thanks) 



4 years ago Comment

2Ca3(PO4)2 + 6SiO2 + 10C ---> P4 + 6CaSiO3 + 10CO

1 mole of Ca3(PO4)2 = 310g
1 mole of P4 = 124g

according to the reaction:
2*310g Ca3(PO4)2----------------124g P4
x g Ca3(PO4)2 ------------------------ 30g P4
x = 150g Ca3(PO4)2

so, your answer is good