Question

The double replacement reaction of aqueous Li3PO4 and CaS produce a calcium phosphate precipitate and aqueous lithium sulfide. How many grams of Ca3(PO4)2 will be produced when 1.13 grams of calcium sulfide react with excess lithium phosphate.

Answer 1

First, use the molar mass of calcium sulfide (72.143 g/mol) to find the number of moles of CaS in 1.13 grams:


`(1.13\ g\ CaS)((mol)/(72.143\ g))=0.01566\ mol\ CaS`

Now the equation for this reaction is
Li3PO4 + CaS --> Ca3(PO4)2 + Li2S

but the BALANCED equation is

2Li3PO4 + 3CaS --> Ca3(PO4)2 + 3Li2S

because in this, there is an equal number of every atom. In the balanced equation, there are 3 moles of CaS for every one mole of Ca3(PO4)2 created, so we change it like this:


`(0.01566\ mol\ CaS)((mol\ Ca_3(PO_4)_2)/(3\ mol\ CaS))=0.00522\ mol\ Ca_3(PO_4)_2`

The molar mass of Ca3(PO4)2 is 310.1767 g/mol:


`(0.00522\ mol\ Ca_3(PO_4)_2)((310.1767\ g)/(mol))=1.619\ g\ Ca_3(PO_4)_2`