Question

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:

Answer 1

\[K_2SO_4(aq)+Pb(C_2H_3O_2)_2(aq)\rightarrow2KC_2H_3O_2(aq)+PbSO_4(s)\]

The solid, \[PbSO_4\] is collected, dried, and found to have a mass of 0.993g. Determine the:

1) Limiting Reactant
2) Theoretical Yield
3) Percent Yield

Answer 2

First, to find the limiting reactant, you need to find the moles of each of your product.

Answer 3

of the desired product

Answer 4

1) Limiting Reactant:

(0.122M)(0.058L) = 0.007076 mol \[K_2SO_4\]
and

(0.102M)(0.04L) = 0.00408 mol \[Pb(C_2H_3O_2)_2\]

\[Pb(C_2H_3O_2)_2\] is the Limiting Reactant.

Answer 5

Yes. The one with the smallest value for the moles is your lmiiting reagent. the one that is consumed thru the rteaction.

Answer 6

How do I calculate the Theoretical Yield and the Percent Yield?

Answer 7

now, find the mass of the product that you should of got - the solid.

Answer 8

To determine theoretical yield, multiply the amount of moles of the limiting reagent by the ratio of the limiting reagent and the synthesized product and by the molecular weight of the product.

Answer 9

2) Theoretical Yield

Like this?

\[0.00408molPb(C_2H_3O_2)_2\times \frac{ 325.288g}{ 1molPb(C_2H_3O_2)_2}\]
\[=1.32717504gPb(C_2H_3O_2)_2\]

Answer 10

oh sorry.

Answer 11

theroetical yield is how much product will SHOULD BE made.

Answer 12

so, that is why i told you to calculate the grams of the desired product!

Answer 13

PERCENT YIELD = \(\sf \color{red}{\frac{actual~yield}{theoretical~yield}}\)

Answer 14

multiplied by 100.

Answer 15

actual yield is how much you got from your experiment.

Answer 16

do you follow me?

Answer 17

So for the Theoretical Yield, all I need to do is to get the grams of both \[2KC_2H_3O_2 \] and \[PbSO_4\] and add them together?

Answer 18

No. The theoretical yield is the product you're trying to make: \(\sf PbSO_4\)

Answer 19

As for the Percent Yield, it is just: (0.993g divided by the "Theoretical Yield") Multiplied by 100?

Answer 20

Ohhhhhh

Answer 21

yes: \(\sf \frac{0.993~g}{\color{red}{theoretical~PbSO_4}}\)

Answer 22

Okay, that's two down! Now I'm still processing the Theoretical Yield part....

Answer 23

theoretical is the amount you were SUPPOSED to get. That means starting with "\(n\)" grams of whaever, you SHOULD of got \(X\) grams of PbSO\(_4\).
But, unfortunately, thats VERY hard to achieve without the use of proper equipment. Ideally, you would want 100% yield. Meaning, all of your product to be \(X\)...but like i said it's hard to do so.

Answer 24

Oh yes, I understand that part.... I'm just thinking about how to start the equation...

Answer 25

So I start with 0.00408 mol \[Pb(C_2H_3O_2)_2\]

and multiply it by the ratio of \[Pb(C_2H_3O_2)_2\] and \[PbSO_4\]

which is 1:1.

Therefore:
\[0.00408molPb(C_2H_3O_2)_2\times \frac{ 1molPbSO_4 }{ 1molPb(C_2H_3O_2)_2 }\]
= 0.00408 mol \[PbSO_4\]
for the first step, yes?

Answer 26

and then, I multiply \[0.00408molPbSO_4\] by the Molecular Weight of \[PbSO_4 \] (which is 303.26g), and get: 1.2373008g.

Answer 27

Okay! I got it! Thank You very much!!!