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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 :

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

[tex](1.13\ g\ CaS)(\frac{mol}{72.143\ g})=0.01566\ mol\ CaS[/tex]

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:

[tex](0.01566\ mol\ CaS)(\frac{mol\ Ca_3(PO_4)_2}{3\ mol\ CaS})=0.00522\ mol\ Ca_3(PO_4)_2[/tex]

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

[tex](0.00522\ mol\ Ca_3(PO_4)_2)(\frac{310.1767\ g}{mol})=1.619\ g\ Ca_3(PO_4)_2[/tex]

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