1. Write the equation and balance it.
3Pb(NO3)2 + 2Na3PO4 ==> Pb3(PO4)2 + 6NaNO3
2. Convert 140 mL of 0.400 M Pb(NO3)2 to moles. mols = M x L.
3. Using the coefficients in the balanced equation, convert moles Pb(NO3)2 from above into moles Na3PO4 used.
4. Now convert moles Na3PO4 to volume.
M = moles/L. You know moles and M, solve for liters.
What volume of 0.300 M Na3PO4 is required to precipitate all the lead(II) ions from 140.0 mL of 0.400 M Pb(NO3)2?
I desperately need help with this, I cannot figure it out, my mind is bleeding.
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Thankyou soooo much! I really appreciate your help
3 answers