Question

10.0 mL of a 0.389 M sodium phosphate solution reacts with 18.9 mL of a 0.167 M lead(II) nitrate solution. What mass of precipitate will form?

Answer 1

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

moles of Na3PO4 = 0.389 x 10/1000 = 0.00389

moles of Pb(NO3)2 = 0.167 x 18.9/1000 = 0.00316

2 moles Na3PO4 reacts with 3 moles Pb(NO3)2

0.00389 moles Na3PO4 reacts with 0.00389 x 3 / 2 = 0.00583 moles Pb(NO3)2

but we have only 0.00316 moles Pb(NO3)2

so Pb(NO3)2 is limiting reagent.

3 moles Pb(NO3)2 forms 811.54 g precipitate

0.00316 moles Pb(NO3)2 forms 0.00316 x 811.54 / 3 = 0.855 g

mass of precipitate formed = 0.855 g

answered by: anonymous