Asked by Andy

15.0 mL of 0.30 M sodium phosphate solution reacts with 20.0 mL of 0.23M lead(II) nitrate solution. What mass of precipitate will form?
Answered by DrBob222
2Na3PO4 + 3Pb(NO3)2 ==> 6NaNO3 + Pb3(PO4)2
mols Na3PO4 = M x L
Using the coefficients in the balanced equation, convert mols Na3PO4 to mols Pb3(PO4)2.

Then convert mols ppt to grams. g = mols x molar mass.
Answered by Shyloh
I don’t kneo
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