# Chemistry lab - limiting reagent - Can someone verify?

• John Ker

## Homework Statement

Suppose that, for reaction 4, you could not find the bottle of 7 M H2SO4 so you added 5.00 mL of the 1.00 M H2SO4 instead. How would this impact your final yield of Copper. (Show with calculations how this would impact the limiting reagent.)

## Homework Equations

CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)

## The Attempt at a Solution

This was the series of transtions done to the copper.
Cu(NO3)2(aq) + 2NaOH(aq) --> Cu(OH)2(s) + 2 NaNO3(aq)

Cu(OH)2(s) + heat --> CuO(s) + H2O(l)

CuO(s) + H2SO4(aq) --> CuSO4(aq) + H2O(l)

CuSO4(aq) + Zn(s)--> ZnSO4(aq) + Cu(s)

I initially started with 10.00ml of .4M Cu(NO3)2. Hence .004 moles. The ratio for each copper reactant to the desired copper product is 1:1. Hence we will end up with .004 CuSO4. Now, in the case of using 7M H2SO4, the CuSO4 is the limiting reagent. However, we use 1M * .005ml = .005mol, doesn't this also keep CuSO4 as the limiting reagent, having no effect on the final yield of copper?

Thanks!

I initially started with 10.00ml of .4M Cu(NO3)2. Hence .004 moles. The ratio for each copper reactant to the desired copper product is 1:1. Hence we will end up with .004 CuSO4.

We end with Cu, not CuSO4, but yes, 1:1 and 0.004 moles is a constant throughout the process (assuming other reagents are in excess).

Now, in the case of using 7M H2SO4, the CuSO4 is the limiting reagent.

No, it is not, limiting reagent is either CuO or H2SO4. Product is never a limiting reagent.

Somehow your final conclusion (no effect of the final yield) is a correct one.

We end with Cu, not CuSO4, but yes, 1:1 and 0.004 moles is a constant throughout the process (assuming other reagents are in excess).

No, it is not, limiting reagent is either CuO or H2SO4. Product is never a limiting reagent.

Somehow your final conclusion (no effect of the final yield) is a correct one.

Hello, thanks for the response.

So after taking a relook youre right, I overlooked the product being the limiting reagent. In that case, wouldn't CuO be the limiting reagent, since it is only .004moles.
CuO is .004
7M H2SO4 is .035M
1M H2SO4 is .005M

In both cases, even with a reduction in the molarity of H2SO4, CuO is still the limiting reagent, hence the same amount of CuSO4 is produced?

In both cases, even with a reduction in the molarity of H2SO4, CuO is still the limiting reagent, hence the same amount of CuSO4 is produced?

In general - yes.

Although high concentration of the acid will definitely speed up the reaction, so while theoretically everything is perfectly OK, process can be difficult to do in practice.