Percent yield crazy--help with calculations?

[alphaj]

Homework Statement

Consider:

2 mol Reactant A (aq) + 3 mol Reactant B (aq) = 6 mol Product A (aq) + 1 mol Product B (s)

You have 0.00075 L of a 0.0514M solution of Reactant A. In the lab, you combine it with 0.042 L of 0.0753M of Reactant B. You get a solid product and an aqueous product (a precipitation reaction!). Please, find the limiting reagent. Then consider this: if you recover 0.0569 g of Product B, then what is your percent yield? Pretend Product B's weight is 310.18g/mol.

Homework Equations

N/a

The Attempt at a Solution

[/B]
Finding the Limiting Reagent:

Considering Reactant A
[(7.5*10^-4 L Reactant A/x moles Reactant A] = [(1 L Reactant A)/0.054 moles Reactant A]= 3.855*10^-5 moles reactant A used

[3.855*10^-5 moles reactant A/x moles Product B]=[2 moles Reactant A/1 mole Product B]= 1.9275*10^-5 moles product B made

Considering Reactant B
[0.042L Reactant B/x moles Reactant B]=[1 L Reactant B/0.0753 moles Reactant B]= 0.0032 moles Reactant B used

[0.0032 moles Reactant B/ x moles Product B]=[3 moles Reactant B/1 mole Product B]= 0.0011 moles Product B made

:.Limiting Reagent is Reactant A.

Yields:
Theoretical yield:
1.9275*10^-5 moles product B *(310.18g/mol)= 0.006 g Product B

Percent Yield: (0.0569g/0.006g)*100 = 948.3%

Well. That doesn't seem right, but I can't find where I went wrong.

 

[mfb]

[(7.5*10^-4 L Reactant A/x moles Reactant A] = [(1 L Reactant A)/0.054 moles Reactant A]= 3.855*10^-5 moles reactant A used

[3.855*10^-5 moles reactant A/x moles Product B]=[2 moles Reactant A/1 mole Product B]= 1.9275*10^-5 moles product B made

That is a very strange notation.

I get the same result as you, I guess the problem statement has a factor 10 wrong somewhere.

You shouldn't round the theoretical yield to one digit, by the way, it introduces a large rounding error.

 

[Borek]

Yep, must be an error in the problem statement.