Question concerning molarity, volume and mass, please help!

Question

If 10 drops of 0.5 mol L-1 Na3PO4(aq) had been added to the sample containing excess 0.25 mol L-1 Ca(NO3)2(aq), what mass of precipitate would you expect to produce? You may assume one drop is equal to a volume of 0.05 mL.

Answer 1

First, write the balanced equation for the reaction between Na3PO4 and Ca(NO3)2

2Na₃PO₄(aq) + 3Ca(NO₃)₂(aq) ==> 6NaNO₃(aq) + Ca₃(PO₄)₂(s)

Next, calculate moles of Na₃PO₄ added

0.05 ml x 1 L/1000 ml x 0.5 moles/L = 0.000025 moles added

Since there is excess Ca(NO₃)₂, the amount of precipitate is dictated and dependent on the moles of N_a3PO₄.  Note the mole ratio of Na₃PO₄ to Ca₃(PO₄)₂ is 2:1.  Using this ratio, calculate moles of Ca₃(PO₄)₃ formed

0.000025 moles Na₃PO₄ x 1 mole Ca₃(PO₄)₂/2 moles Na₃PO₄ = 0.0000125 moles Ca₃(PO₄)₂ formed

Convert moles to mass (g)

0.0000125 moles Ca₃(PO₄)₂ x 310 g/mole = 0.003875 g = 3.875 mg = 4 mg (to 1 significant figure)