A57.0 ml sample of a 0.120 m potassium sulfate solution is mixed with 35.5 ml of a 0.118 m 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 0.992 g . determine the limiting reactant, the theoretical yield, and the percent yield.

Limiting reagent = lead(II) acetate

Theoretical yield = 1.2704 g

% yield = 78.09 %

Explanation:

Considering:

Or,

Given :

For potassium sulfate :

Molarity = 0.120 M

Volume = 57.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 57.0×10⁻³ L

Thus, moles of potassium sulfate:

Moles of potassium sulfate  = 0.00684 moles

For lead(II) acetate :

Molarity = 0.118 M

Volume = 35.5 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 35.5×10⁻³ L

Thus, moles of lead(II) acetate :

Moles of lead(II) acetate  = 0.004189 moles

According to the given reaction:

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

0.00684 moles potassium sulfate react with 0.00684 mole of lead(II) acetate

Moles of lead(II) acetate = 0.004189 moles

Limiting reagent is the one which is present in small amount. Thus, lead(II) acetate is limiting reagent. ( 0.004189 < 0.00684)

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

0.004189 mole of lead(II) acetate gives 0.004189 mole of lead(II) sulfate

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

Mass of lead(II) sulfate = Moles × Molar mass = 0.004189 × 303.26 g = 1.2704 g

Theoretical yield = 1.2704 g

Given experimental yield = 0.992 g

% yield = (Experimental yield / Theoretical yield) × 100 = (0.992/1.2704 g) × 100 = 78.09 %

yes definetly because you are using energy