If a solution containing 19.35 g of lead(II) nitrate is allowed to react completely with a solution containing 6.256 g of sodium sulfate, how many grams of solid precipitate will be formed?

Question

How many grams of the reactant in excess will remain after the reaction?

Assuming complete precipitation, how many moles of each ion remain in the solution? If an ion is no longer in solution, you should enter in a zero for the number of moles.

Answer 1

The reaction between lead(II) nitrate and sodium sulfate is a precipitation reaction that can be represented by the following balanced chemical equation:

Pb(NO3)2 (aq) + Na2SO4 (aq) -> PbSO4 (s) + 2NaNO3 (aq)

From this equation, we can see that 1 mole of lead(II) nitrate reacts with 1 mole of sodium sulfate to produce 1 mole of lead(II) sulfate precipitate.

First, we need to calculate the number of moles of each reactant. The molar mass of Pb(NO3)2 is approximately 331.21 g/mol, and the molar mass of Na2SO4 is approximately 142.04 g/mol.

Number of moles of Pb(NO3)2 = 19.35 g / 331.21 g/mol = 0.058 mol

Number of moles of Na2SO4 = 6.256 g / 142.04 g/mol = 0.044 mol

The limiting reactant is the one with the smallest number of moles, which in this case is Na2SO4, because we have less of it in terms of moles. This will determine the maximum amount of PbSO4 that can be produced.

From the balanced chemical equation, we see that 1 mole of Na2SO4 produces 1 mole of PbSO4. Therefore, 0.044 mol of Na2SO4 will produce 0.044 mol of PbSO4.

To find the mass of the PbSO4 precipitate, we multiply the number of moles by its molar mass (approximately 303.27 g/mol):

Mass of PbSO4 = 0.044 mol * 303.27 g/mol = 13.34 g

So, 13.34 g of lead(II) sulfate precipitate will be formed.

Answer 2

First, let's start to understand the problem by making an equation.

1. Lead (II) nitrate is Pb(NO₃)₂ since lead has a +2 charge as given and nitrate has a -1 charge.
2. Sodium sulfate is Na₂SO₄ since sodium has a +1 charge and sulfate has a -2 charge.

If we balance our equation, we get:

Pb(NO₃)₂ (aq) + Na₂SO₄ (aq) --> PbSO₄ (s) + 2 NaNO₃ (aq)

Now, we need to find the limiting reactant. The molar mass of Pb(NO₃)₂ is 331.2 g/mol, the molar mass of Na₂SO₄ is 142.04 g/mol, and the molar mass of PbSO₄ is 303.26 g/mol.

If all of the 19.35g of lead (II) nitrate reacts, that is 19.35÷331.2 mol lead (II) nitrate, which is 0.0584 mol lead (II) nitrate. That means if all of the lead (II) nitrate reacts, the reaction would also need 0.0584 mol sodium sulfate. That is 0.0584(142.04), or ~8.295g sodium sulfate.

We only have 6.256g sodium sulfate, which is <8.295. Thus, there's not enough sodium sulfate for the lead (II) nitrate to fully react, which makes sodium sulfate the limiting reactant.

Let's answer the first question: how much solid precipitate (i.e. PbSO₄) will be formed?

Since sodium sulfate is the limiting reactant, the amount of mol of sodium sulfate will be the amount of mol of lead (II) nitrate that reacts. Since there's a 1:1:1 ratio of mol sodium sulfate, mol lead (II) nitrate, and mol PbSO₄, the answer to the first question is mol sodium sulfate (which is also mol PbSO₄) multiplied by the molar mass of PbSO₄.

grams PbSO₄ = mol PbSO₄ * 303.26 = (6.256 mol Na₂SO₄ ÷ 142.04 g/mol Na₂SO₄)(303.26 g/mol PbSO₄) = (6.256/142.04)(303.26) = 13.357g PbSO₄, or 13.357g solid precipitate.

Let's answer the second question: how much of reactant in excess will remain after the reaction?

Since the excess reactant is lead (II) nitrate, we need to first find how much lead (II) nitrate was reacted. Since 0.044 mol of lead (II) nitrate was reacted, which was found by dividing 6.256 grams of sodium sulfate with it's molar mass of 142.04 g/mol, that means the amount of lead (II) nitrate reacted was:

(0.044 mol Pb(NO₃)₂)(331.2 g/mol Pb(NO₃)₂) = 0.044(331.2) = 14.573g lead (II) nitrate reacted

It's asking for the leftover reactant, which would be the total amount subtracted by the amount reacted.

(19.35g lead (II) nitrate total) - (14.573g lead (II) nitrate reacted) = 19.35 - 14.573 = 4.777g excess reactant

Let's answer the third question: how many moles of each ion remain in the solution?

First, we can identify which ions are remaining in the solution. We have 4 ions: Pb2+, Na+, NO₃-, and SO₄2-. Since all of the sodium sulfate reacted, all of the SO₄2- became part of the precipitate, so there's no more of that ion in the solution. However, since not all of the Pb2+ and NO₃- reacted, both of those ions are still in the aqueous solution. Also, because the Na+ ion became part of the aqueous NaNO₃, both ions are present in the solution.

To find the mol Pb2+ that wasn't reacted, we do:

(0.0584 mol Pb(NO₃)₂ total) - (0.0440 mol Pb(NO₃)₃ reacted) = 0.0584-0.044 = 0.0144 mol unreacted Pb(NO₃)₂

Because there's 1 mol of Pb2+ for every mol Pb(NO₃)₂, there is a final 0.0144 mol Pb2+ in the solution.

Now, regarding NO₃-, we know that all of the ion became part of the aqueous solution. Since there was 0.0584 mol Pb(NO₃)₂ and there's 2 mol NO₃- for every mol Pb(NO₃)₂, there is a final 2(0.0584) = 0.1168 mol NO₃- in the solution.

Now, regarding Na+, we know (just like NO₃-) that all of the ion became part of the aqueous solution. Since there was 0.044 mol of Na₂SO₄, and there's 2 mol Na+ for every mol Na₂SO₄, there is a final 2(0.044) = 0.088 mol Na+ in the solution.

Just to summarize the answer to the third question:

0.0144 mol Pb2+

0.1168 mol NO₃-

0.088 mol Na+

0 mol SO₄

I hope this helped you out, I tried to be thorough with my explanations. Let me know if you need any clarification!!