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

The percent yield of the reaction is 80%

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%

i think 4.44 moles

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.