Question

A 56.0 mL sample of a 0.116 M potassium sulfate solution is mixed with 38.5 mL of a 0.102 M lead(II) acetate solution and

Answer 1

The limiting reactant is lead(II) acetate the theoretical yield is 1.1909 g and the percent yield is 83.71 %.

What is a molarity ?

The amount of the sample in a specific solution volume is known as its molarity (M). The molarity of a solute per liter of the a solution is known as molar ratio. The molar concentration in a solution is another name for molarity.

Briefing:

Molarity

= moles of solute/volume of the solution

or Moles

= Molarity * volume of solution

For potassium sulfate :

Molarity = 0.116 M

Volume = 56.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 56.0×10⁻³ L

Thus, moles of potassium sulfate:

Moles = 0.116 M * 56.0 * 10-3 moles

Moles of potassium sulfate  = 0.006496 moles

For lead(II) acetate :

Molarity = 0.102 M

Volume = 38.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.5×10⁻³ L

Thus, moles of lead(II) acetate :

Moles = 0.102 * 38.5 *10-3 moles

Moles of lead(II) acetate  = 0.003927 moles

According to the given reaction:

$\mathrm{K}_2 \mathrm{SO}_{4(a q)}+\mathrm{Pb}\left(\mathrm{C}_2 \mathrm{H}_3 \mathrm{O}_2\right)_{2(a q)} \rightarrow 2 \mathrm{KC}_2 \mathrm{H}_3 \mathrm{O}_{2(s)}+\mathrm{PbSO}_{4(a q)}$

1 mole of potassium sulfate react with 1 mole of lead(II) acetate

0.006496 moles potassium sulfate react with 0.006496 mole of lead(II) acetate

Moles of lead(II) acetate = 0.003927 moles

Lead(II) acetate is limiting reagent. ( 0.003927 < 0.006496)

0.003927 mole of lead(II) acetate gives 0.003927 mole of lead(II) sulfate

Molar mass of lead(II) sulfate = 303.26 g/mol

Mass of lead(II) sulfate = Moles × Molar mass = 0.003927 × 303.26 g = 1.1909 g

Theoretical yield = 1.1909 g

Given experimental yield = 0.997 g

% yield = (Experimental yield / Theoretical yield) × 100 = (0.997/1.1909 g) × 100 = 83.71 %

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