Respuesta :
Answer:
Lead(II) acetate is limiting reagent. (0.00364 < 0.006832)
Theoretical yield = 1.104 g
% yield = (Experimental yield / Theoretical yield) × 100 = (0.997 / 1.104) × 100 = 90.31 %
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For potassium sulfate :
Molarity = 0.112 M
Volume = 61.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 61.0×10⁻³ L
Thus, moles of potassium sulfate :
[tex]Moles=0.112 \times {61.0\times 10^{-3}}\ moles[/tex]
Moles of potassium sulfate = 0.006832 moles
For lead(II) acetate :
Molarity = 0.104 M
Volume = 35.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 35.0×10⁻³ L
Thus, moles of lead(II) acetate :
[tex]Moles=0.104 \times {35.0\times 10^{-3}}\ moles[/tex]
Moles of lead(II) acetate = 0.00364 moles
According to the given reaction:
[tex]KSO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow PbSO_4_{(s)}+2KC_2H_3O_2_{(aq)}[/tex]
1 mole of potassium sulfate reacts with 1 mole of lead(II) acetate
So,
0.006832 mole of potassium sulfate reacts with 0.006832 mole of lead(II) acetate
Moles of lead(II) acetate = 0.006832 mole
Available moles of lead(II) acetate = 0.00364 mole
Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. (0.00364 < 0.006832)
The formation of the product is governed by the limiting reagent. So,
1 mole of lead(II) acetate gives 1 mole of lead(II) sulfate
0.00364 mole of lead(II) acetate gives 0.00364 mole of lead(II) sulfate
Moles of lead(II) sulfate = 0.00364 moles
Molar mass of lead(II) sulfate = 303.26 g/mol
Mass of lead(II) chloride = Moles × Molar mass = 0.00364 × 303.26 g = 1.104 g
Theoretical yield = 1.104 g
Given experimental yield = 0.997 g
% yield = (Experimental yield / Theoretical yield) × 100 = (0.997 / 1.104) × 100 = 90.31 %
