Chemistry, 05.11.2019 06:31 Esan2435

A61.0ml sample of a 0.112m potassium sulfate solution is mixed with 35.0ml of a 0.104m lead(ii) acetate solution and the following precipitation reaction occurs:
k2so4(aq)+pb(c2h3o2)2(aq)? 2kc2h3o2(aq)+pbso4(s)
the solid pbso4 is collected, dried, and found to have a mass of 0.997g .
determine the limiting reactant, the theoretical yield, and the percent yield.
part a.
identify the limiting reactant.
kc2h3o2
pb(c2h3o2)2
k2so4
pbso4
part b.
determine the theoretical yield.
mass of pbso4 =
part c.
determine the percent yield=

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Answer from: lalacada1

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:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

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 :

$Moles=0.112 \times {61.0\times 10^{-3}}\ moles$

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 :

$Moles=0.104 \times {35.0\times 10^{-3}}\ moles$

Moles of lead(II) acetate = 0.00364 moles

According to the given reaction:

$KSO_4_{(aq)}+Pb(C_2H_3O_2)_2_{(aq)}\rightarrow PbSO_4_{(s)}+2KC_2H_3O_2_{(aq)}$

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 %

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A61.0ml sample of a 0.112m potassium sulfate solution is mixed with 35.0ml of a 0.104m lead(ii) acet...

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