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?
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.
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.
To find out the volume of 0.300 M Na3PO4 required to precipitate all the lead(II) ions from 130.0 mL of 0.400 M Pb(NO3)2, we need to use the concept of stoichiometry.
Step 1: Write the balanced chemical equation.
The balanced chemical equation for the reaction between Na3PO4 and Pb(NO3)2 is:
3Na3PO4 + 2Pb(NO3)2 → Pb3(PO4)2 + 6NaNO3
Step 2: Determine the moles of Pb(NO3)2.
We can use the concentration and volume of Pb(NO3)2 to calculate the number of moles:
moles of Pb(NO3)2 = concentration × volume
moles of Pb(NO3)2 = 0.400 M × 130.0 mL
Note: It is important to convert the volume from mL to liters by dividing by 1000.
Step 3: Use stoichiometry to find moles of Na3PO4.
Since the balanced equation shows a 3:2 ratio between Na3PO4 and Pb(NO3)2, we can use this ratio to find the moles of Na3PO4 needed:
moles of Na3PO4 = (moles of Pb(NO3)2) × (3 moles of Na3PO4 / 2 moles of Pb(NO3)2)
Step 4: Calculate the volume of Na3PO4.
To calculate the volume, we need to know the concentration of Na3PO4. Given that it is 0.300 M:
volume of Na3PO4 = (moles of Na3PO4) / (concentration of Na3PO4)
By following these steps and performing the calculations, you can find the volume of 0.300 M Na3PO4 required to precipitate all the lead(II) ions from 130.0 mL of 0.400 M Pb(NO3)2.