Question

When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.

Answer 1

Answer:

The mass of PbSO4 formed 15.163 gram

Explanation:

mole of Pb(NO₃)₂ = 1.25 x 0.05 = 0.0625

mole of Na₂SO₄ = 2 x 0.025 = 0.05

                                      Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

( Mole/Stoichiometry )    $\frac{0.0625}{1}$           $\frac{0.05}{1}$

                                     = 0.0625     = 0.05

From  (Mole/ Stoichiometry ) we can conclude that Na₂SO₄ is limiting reagent.

Mass of PbSO₄ precipitate = 0.05 x Molecular mass of PbSO₄

                                            = 0.05 x 303.26 g

                                            = 15.163 g

Answer 2

The mass of PbSO₄ obtained from the reaction is 15.15 g

We'll begin by calculating the number of mole of Pb(NO₃)₂ and Na₂SO₄ in the solution.

For Pb(NO₃)₂:

Volume = 1.25 L

Molarity = 0.05 M

Mole of Pb(NO₃)₂ =?

Mole = Molarity × Volume

Mole of Pb(NO₃)₂ = 0.05 × 1.25

Mole of Pb(NO₃)₂ = 0.0625 mole

For Na₂SO₄:

Volume = 2 L

Molarity = 0.025 M

Mole of Na₂SO₄ =?

Mole = Molarity × Volume

Mole of Na₂SO₄ = 0.025 × 2

Mole of Na₂SO₄ = 0.05 mole

• Next, we shall determine the limiting reactant.

Pb(NO₃)₂ + Na₂SO₄ → PbSO₄ + 2 NaNO₃

1 mole : 1 mole

0.0625 mole : 0.05 mole

Thus, the limiting reactant is Na₂SO₄ since little amount is present.

• Next, we shall determine the mole of PbSO₄ produced from the reaction.

From the balanced equation above,

1 mole of Na₂SO₄ reacted to produce 1 mole PbSO₄

Therefore,

0.05 mole of Na₂SO₄ will also react to produce 0.05 mole of PbSO₄

• Finally, we shall determine the mass of 0.05 mole of PbSO₄

Mole of PbSO₄ = 0.05

Molar mass of PbSO₄ = 207 + 32 + (16×4) = 303 g/mol

Mass of PbSO₄ =?

Mass = mole × molar mass

Mass of PbSO₄ = 0.05 × 303

Mass of PbSO₄ = 15.15 g

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