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]
