Respuesta :
Answer: Mass of [tex]PbSO_{4}[/tex] produced is 99.1 g.
Explanation:
Chemical equation for the given reaction is as follows.
[tex]K_{2}SO_{4}(aq) + Pb(NO_{3})_{2}(aq) \rightarrow 2KNO_{3}(aq) + PbSO_{4}(s)[/tex]
Moles of [tex]Pb(NO_{3})_{2}[/tex] = Molarity × volume
= [tex]0.165 M \times 1.98 ml[/tex]
= 0.3267 mol
Moles of [tex]K_{2}SO_{4} = 0.702 M \times 3.6 ml[/tex]
= 2.527 mol
As moles of [tex]Pb(NO_{3})_{2}[/tex] is less. Hence, it is the limiting reactant.
We know that, molar mass of [tex]PbSO_{4}[/tex] is 303.26 g/mol.
Then,
0.3267 mol of [tex]PbSO_{4}[/tex] = [tex]0.3267 \times 303.26[/tex]
= 99.075 g
or, = 99.1 g
Therefore, mass of [tex]PbSO_{4}[/tex] produced is 99.1 g.
The mass of PbSO₄ produced when 1.98 mL of 0.165 M Pb(NO₃)₂ and 3.6 mL of 0.702 M K₂SO₄ are mixed is 0.10 g
We'll begin by calculating the number of mole of Pb(NO₃)₂ and K₂SO₄ in the solution. This is illustrated below:
For Pb(NO₃)₂:
Volume = 1.98 mL = 1.98 / 1000 = 0.00198 L
Molarity = 0.165 M
Mole of Pb(NO₃)₂ =?
Mole = Molarity x Volume
Mole of Pb(NO₃)₂ = 0.165 × 0.00198
Mole of Pb(NO₃)₂ = 0.0003267 mole
For K₂SO₄:
Volume = 3.6 mL = 3.6 / 1000 = 0.0036 L
Molarity = 0.702 M
Mole of K₂SO₄ =?
Mole = Molarity x Volume
Mole of K₂SO₄ = 0.702 × 0.0036
Mole of K₂SO₄ = 0.0025272 mole
Next, we shall determine the limiting reactant. This can be obtained as follow:
K₂SO₄ + Pb(NO₃)₂ —> PbSO₄ + 2KNO₃
1 : 1
0.0025272 : 0.0003267
From the above, we can see that the mole of Pb(NO₃)₂ is less. Thus, Pb(NO₃)₂ is the limit reactant.
Next, we shall determine the number of mole of PbSO₄ produced from the reaction. This can be obtained as follow:
From the balanced equation above,
1 mole of Pb(NO₃)₂ reacted to produce 1 mole of PbSO₄.
Therefore,
0.0003267 mole of Pb(NO₃)₂ will also react to produce 0.0003267 mole of PbSO₄.
Finally, we shall determine the mass of 0.0003267 mole of PbSO₄.
Mole of PbSO₄ = 0.0003267 mole
Molar mass of PbSO₄ = 207 + 32 + (16×4)
= 303 g/mol
Mass of PbSO₄ =?
Mass = mole × molar mass
Mass of PbSO₄ = 0.0003267 × 303
Mass of PbSO₄ = 0.0989901 g
Mass of PbSO₄ ≈ 0.10 g (3 sf)
Thus, the mass of PbSO₄ produced from the reaction is 0.10 g
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