Question

11.4 mL of a 0.343 M sodium phosphate solution reacts with 16.9 mL of a 0.174 M lead(II) nitrate solution. What mass of precipitate will form?

Answer 1

The balanced equation is

2 Na₃PO₄(aq) + 3 Pb(NO₃)₂ (aq) ------> Pb₃(PO₄)₂ (s) + 6 NaNO₃(aq)

Number of moles of sodium phosphate = molarity * volume of solution in L

Number of moles of sodium phosphate = 0.343 * 0.0114 L = 0.00391 mol

Number of moles of lead (II) nitrate = 0.174 * 0.0169 L = 0.00294 mol

From the balanced equation we can say that

2 mole of Na3PO4 requires 3 mole of Pb(NO3)2 so

0.00391 mole of Na3PO4 will require

= 0.00391 mole of Na3PO4 *(3 mole of Pb(NO3)2 / 2 mole of Na3PO4)

= 0.00587 mole of Pb(NO3)2

But we have 0.00294 mole of Pb(NO3)2 which is in short so Pb(NO3)2 is limiting reactant

From the balanced equation we can say that

3 mole of Pb(NO3)2 produces 1 mole of Pb3(PO4)2 so

0.00294 mole of Pb(NO3)2 will produce

= 0.00294 mole of Pb(NO3)2 *(1 mole of Pb3(PO4)2 / 3 mole of Pb(NO3)2)

= 0.000980 mole of Pb3(PO4)2

mass of 1 mole of Pb3(PO4)2 = 811.5427 g

so the mass of 0.000980 mole of Pb3(PO4)2 = 0.795 g

Therefore, the mass of Pb3(PO4)2 produced would be 0.795 g