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.
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.
Thankyou soooo much! I really appreciate your help
To determine the volume of 0.300 M Na3PO4 required to precipitate all the lead(II) ions from 140.0 mL of 0.400 M Pb(NO3)2, we first need to determine the stoichiometry of the reaction between Na3PO4 and Pb(NO3)2.
The balanced equation for the reaction is:
3Pb(NO3)2 + 2Na3PO4 -> Pb3(PO4)2 + 6NaNO3
From the equation, we can see that 3 moles of Pb(NO3)2 react with 2 moles of Na3PO4 to form 1 mole of Pb3(PO4)2.
Step 1: Calculate the moles of Pb(NO3)2 in solution.
Moles of Pb(NO3)2 = concentration of Pb(NO3)2 × volume of Pb(NO3)2
= 0.400 M × 140.0 mL
= 0.400 mol/L × 0.140 L
= 0.056 mol
Step 2: Determine the moles of Na3PO4 required.
From the stoichiometry of the reaction, we can see that 3 moles of Pb(NO3)2 react with 2 moles of Na3PO4. Therefore, the moles of Na3PO4 required can be calculated using the ratio:
Moles of Na3PO4 = (moles of Pb(NO3)2 × 2) / 3
= (0.056 mol × 2) / 3
≈ 0.037 mol
Step 3: Calculate the volume of 0.300 M Na3PO4 required.
Volume of Na3PO4 = Moles of Na3PO4 / Concentration of Na3PO4
= 0.037 mol / 0.300 M
≈ 0.123 L or 123 mL
Therefore, approximately 123 mL 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.
To solve this problem, we need to use the concept of stoichiometry and the balanced chemical equation for the reaction. In this case, the balanced chemical equation is:
2 Na3PO4(aq) + 3 Pb(NO3)2(aq) -> Pb3(PO4)2(s) + 6 NaNO3(aq)
Let's break down the steps to solve this problem:
1. Identify and balance the equation: We have already provided the balanced equation above.
2. Determine the stoichiometric ratio: From the balanced equation, we see that 2 moles of Na3PO4 react with 3 moles of Pb(NO3)2 to produce 1 mole of Pb3(PO4)2. This means that the stoichiometric ratio is 2:3:1 for Na3PO4:Pb(NO3)2:Pb3(PO4)2.
3. Calculate the number of moles of Pb(NO3)2: To do this, we use the given concentration and volume of Pb(NO3)2.
Moles of Pb(NO3)2 = (concentration of Pb(NO3)2) × (volume of Pb(NO3)2 in liters)
= 0.400 mol/L × (0.1400 L)
= 0.0560 mol
4. Determine the number of moles of Na3PO4 required: Using the stoichiometric ratio, we can find the number of moles of Na3PO4 required.
Moles of Na3PO4 = (moles of Pb(NO3)2) × (2/3)
= 0.0560 mol × (2/3)
= 0.0373 mol
5. Calculate the volume of 0.300 M Na3PO4 required: Now we can determine the volume of 0.300 M Na3PO4 solution needed.
Volume of Na3PO4 = (moles of Na3PO4) / (concentration of Na3PO4)
= 0.0373 mol / 0.300 mol/L
= 0.124 liters or 124 mL
Therefore, the volume of 0.300 M Na3PO4 required to precipitate all the lead(II) ions from 140.0 mL of 0.400 M Pb(NO3)2 is 124 mL.