Question

Calcium oxide (CaO) reacts with tetraphosphorus decoxide (P4O10) to form calcium phosphate (Ca3(P4O10)2) according to this equation: 6 CaO+P4O10→2Ca3(PO4)2 In a particular reaction, 14.2 g P4O10 reacts with excess CaO and produces 24.8 g Ca3(PO4)2 . Approximately what is the percent yield of the reaction

Answer 1

The percent yield of the reaction is 80%

Step-by-step explanation:

6CaO + P4O10→2Ca3(PO4)2

Relation is 1 mol P4O10 to make 2 moles Ca3(PO4)2 (100% yield)

Molar mass P4O10 = 283,8 g/m

Molar mass Ca3(PO4)2 = 309,8 g/m

Mass /Molar mass = moles

I have 14,2 g P4O10 so I have 14,2g /283,8g/m = 0,05 moles

It was converted 24,8 g Ca3(PO4)2 so it was converted:

24,8 g /309,8 g/m = 0,08 moles

In the 100 % yield of the reaction:

1 mol P4O10 _____ 2 moles Ca3(PO4)2

I use 0,05 moles P4O10 ____ so I made (0,05 moles . 2 moles)/1 mol = 0,1 moles

0,1 moles converted of Ca3(PO4)2 ____ 100 %

0,08 moles moles I converted of Ca3(PO4)2 ____

(0,08 moles . 100%) / 0,1 moles = 80%