A 58.0mL sample of a 0.122M potassium sulfate solution is mixed with 40.0mL of a 0.102M Lead (II) acetate solution and this 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 0.993g.

Determine the:

1) Limiting Reactant

2) Theoretical Yield

3) Percent Yield

Question

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Chemistry

2 October, 06:59

A 58.0mL sample of a 0.122M potassium sulfate solution is mixed with 40.0mL of a 0.102M Lead (II) acetate solution and this 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 0.993g.

Determine the:

1) Limiting Reactant

2) Theoretical Yield

3) Percent Yield

+3

Answers (1)

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Beard · Answer 1

First we need to find the limiting reactant. This is the reactant that is theoretically consumed completely first. Theoretical yield is how much product is produced given the limiting reactant is consumed and the percent yield is the fraction of the actual products produced and the theoretical products produced.

The limiting reactant is Pb (C2 H3 O2) 2.

The theoretical yield is 1.2373008g and you should get the Percent yield which is Actual yield/theoretical yield.