3Pb(NO3)2 + 2Na3PO4 ==> Pb3(PO4)2 + 6NaNO3
mols Pb = M x L = ?
Convert mols Pb to mols Na3PO4 using the coefficients.
Then M Na3PO4 = mols Na3PO4/L Na3PO4. you know mols and M, solve for L.
What volume of 0.300 M Na3PO4 is required to precipitate all the lead(II) ions from 130.0 mL of 0.400 M Pb(NO3)2?
3 answers
I have 0.400*0.13=5.2 so far
I don't believe we're working the same problem. 0.400 x 0.13 = 0.052 mols. That's my step 1.
My step 2. Now use the coefficients in the balanced equation (I balanced the equation for you) to convert mols Pb(NO3)2 to mols Na3PO4.
My step 3.
M Na3PO4 = mols Na3PO4/LNa3PO4. You have mols and M, solve for L.
My step 2. Now use the coefficients in the balanced equation (I balanced the equation for you) to convert mols Pb(NO3)2 to mols Na3PO4.
My step 3.
M Na3PO4 = mols Na3PO4/LNa3PO4. You have mols and M, solve for L.