A 56.0mL sample of a 0.102 M potassium sulfate solution is mixed with 34.7mL of a 0.114 M lead acetate solution and this precipitation reaction occurs:

Question

K2SO4(aq)+Pb(C2H3O2)2(aq)→PbSO4(s)+2KC2H3O2(aq) The solid PbSO4 is collected, dried, and found to have a mass of 1.16 g.

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Determine the theoretical yield.

Determine the percent yield.

Answer 1

find the moles you started with K2SO4, nad moles started with Pbacetate.

the mole ratio in the equation is 1:1, so which ever mole started with is lowest, that is the limiting reactant., and you will get the same number of moles of lead sufate. Change that to grams, that is the theoritical yield

percent yeld= 1.16/theoyield * 100

Answer 2

Is this correct?
K2SO4 0.00573
Pbacetate 0.00307
limiting reactant is Pb(C2H3O2)2.
Grams lead sufate. 303.26g
Is theoretical yield 303.26g?

Percent Yield= 1.16g/303.26g * 100= 38 %

Answer 3

Some parts right; some wrong.
I didn't get the same answer you did when multiplying M x L = ?

For mols K2SO4 I obtained 0.00571 and for
Pb(ac)2 I obtained 0.00396. That doesn't change the identification of the limiting reagent but it points up a problem on using your calculator.
Then for grams PbSO4 that 303.25 is the molar mass. You find grams by mols x molar mass. mols = 0.00571
0.00571 x 303.25 = about 1.7 (you need it more accurately) and that is the theoretical yield.
Then % yield = (1.16/1.7)*100 = ?
While I'm at it let me point out another calculator problem. (1.16/303.26)*100 = 0.38% and not 38%.

Answer 4

Thats wrong!

The correct answer:

Determine the theoretical yield.
1.195 g
Determine the percent yield.
97.0 %