Question

If 100.0 mL of 0.453 M Na2SO4 are added to 100.0 mL of 0.907 M Pb(NO3)2, how many grams of PbSO4 can be produced?
Na2SO4(aq)+Pb(NO3)2(aq)⟶2NaNO3(aq)+PbSO4(s)

Answer 1

Answer: The mass of $PbSO_4$ produced will be 13.73 grams.

Explanation:

To calculate the number of moles from the given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

• For $Na_2SO_4$

Molarity of the solution = 0.453 mol/L

Volume of solution = 100 mL = 0.1 L     (Conversion factor:  1L = 1000 mL)

Putting values in equation 1, we get:

$0.453mol/L=\frac{\text{Moles of }Na_2SO_4}{0.1L}\\\\\text{Moles of }Na_2SO_4=0.0453mol$

• For $Pb(NO_3)_2$

Molarity of the solution = 0.907 mol/L

Volume of solution = 100 mL = 0.1 L

Putting values in equation 1, we get:

$0.907mol/L=\frac{\text{Moles of }Pb(NO_3)_2}{0.1L}\\\\\text{Moles of }Pb(NO_3)_2=0.0907mol$

For the given chemical reaction:

$Na_2SO_4(aq.)+Pb(NO_3)_2(aq.)\rightarrow 2NaNO_3(aq.)+PbSO_4(s)$

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of lead nitrate.

So, 0.0453 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.0453=0.0453moles$ of lead nitrate.

As, the given amount of lead nitrate is more than the required amount. Thus, it is considered as an excess reagent.

Hence, sodium sulfate is considered as an limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of sodium sulfate reacts with 1 mole of lead sulfate.

So, 0.0453 moles of sodium sulfate will react with = $\frac{1}{1}\times 0.0453=0.0453moles$ of lead sulfate.

To calculate the mass of lead sulfate, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Moles of lead sulfate = 0.0453 mol

Molar mass of lead sulfate = 303.26 g/mol

Putting values in above equation, we get:

$0.0453mol=\frac{\text{Mass of lead sulfate}}{303.26g/mol}\\\\\text{Mass of lead sulfate}=13.73g$

Hence, the mass of $PbSO_4$ produced will be 13.73 grams.

Answer 2

The mass of PbSO₄ that could be produced is 13.74 g

From the question,

We are to determine the mass of PbSO₄ that could be produced from the reaction

The given balanced chemical equation for the reaction is

Na₂SO₄(aq) + Pb(NO₃)₂(aq) ⟶ 2NaNO₃(aq) + PbSO₄(s)

This means

1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 2 moles of NaNO₃ and 1 mole of PbSO₄

Now, we will determine the number of moles of each reactant present

• For Na₂SO₄

Volume = 100.0 mL = 0.1 L

Concentration = 0.453 M  

Using the formula

Number of moles = Concentration × Volume

∴ Number of moles of Na₂SO₄ = 0.453 × 0.1

Number of moles of Na₂SO₄ = 0.0453 mol

• For Pb(NO₃)₂

Volume = 100.0 mL = 0.1 L

Concentration = 0.907 M

∴ Number of moles of Pb(NO₃)₂ = 0.907 × 0.1

Number of moles of Pb(NO₃)₂ = 0.0907 mol

Since, 1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂

Then,

0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂

Now, from the balanced chemical equation

1 mole of Na₂SO₄ reacts with 1 mole of Pb(NO₃)₂ to produce 1 mole of PbSO₄

Then,

0.0453 mole of Na₂SO₄ will react with 0.0453 mole of Pb(NO₃)₂ to produce 0.0453 mole of PbSO₄

∴ The number of moles of PbSO₄ produced is 0.0453 mole

Now, for the mass of PbSO₄ produced

From the formula

Mass = Number of moles × Molar mass

Molar mass of PbSO₄ = 303.26 g/mol

∴ Mass of PbSO₄ produced = 0.0453 × 303.26

Mass of PbSO₄ produced = 13.737678 g

Mass of PbSO₄ produced ≅ 13.74 g

Hence, the mass of PbSO₄ that could be produced is 13.74 g

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