What volume of 0.150 M Na_3PO_4 is required to precipitate all the lead(II) ions from 120.0 mL of 0.350 M Pb(NO_3)_2?
2Na3PO4 + 3Pb(NO3)2 ==> Pb3(PO4)2 + 6NaNO3
Just follow the steps in this example of a stoichiometry problem.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the volume of 0.150 M Na3PO4 needed to precipitate all the lead(II) ions from 120.0 mL of 0.350 M Pb(NO3)2, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2 is:
3 Na3PO4 + 2 Pb(NO3)2 -> Pb3(PO4)2 + 6 NaNO3
From the equation, we can see that 3 moles of Na3PO4 reacts with 2 moles of Pb(NO3)2 to produce 1 mole of Pb3(PO4)2.
Step 1: Find the number of moles of Pb(NO3)2.
Given:
Volume of Pb(NO3)2 = 120.0 mL
Molarity of Pb(NO3)2 = 0.350 M
Using the formula:
moles = volume (L) x concentration (M)
Converting mL to L:
Volume (L) = 120.0 mL / 1000 mL/L = 0.120 L
moles of Pb(NO3)2 = 0.120 L x 0.350 M = 0.042 moles
Step 2: Use stoichiometry to find the number of moles of Na3PO4 required.
From the balanced chemical equation, we can see that it takes 2 moles of Pb(NO3)2 to react with 3 moles of Na3PO4.
Therefore, the number of moles of Na3PO4 required = (3/2) x moles of Pb(NO3)2
= (3/2) x 0.042 moles
= 0.063 moles
Step 3: Find the volume of 0.150 M Na3PO4 required.
Molarity (M) = moles / volume (L)
To find the volume of 0.150 M Na3PO4 required, rearrange the equation:
volume (L) = moles / Molarity (M)
= 0.063 moles / 0.150 M
= 0.42 L
Converting L to mL:
Volume (mL) = 0.42 L x 1000 mL/L = 420 mL
Therefore, the volume of 0.150 M Na3PO4 required to precipitate all the lead(II) ions from 120.0 mL of 0.350 M Pb(NO3)2 is 420 mL.