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) 

(1) Answers

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

Add answer