# Thread: I need the solution for this question, i dont get it

1. This is for a PreLab

Question is, Calculate the volume of the 1 mol/L copper(II) sulfate(CuSO4) solution required to react with 1.00g of iron fillings.Use the equation that will give you the most copper(II) sulfate.Add an addittional 10% to your calculated volume in order to have sufficient copper (II) sulfate to ensure an excess.

Use this equation

2Fe + 3Cu :rarrow:2Fe + 3Cu  2.

3. A few questions first: Can u clarify the equation given? And what do u mean when u say Use the equation that will give u the most copper sulphate?

Anyways, if i follow the instructions and equations given exactly, this would be how its done:

step 1) Calculate the no. of mol of Fe in 1.00g
- 1/55.8 = 0.0179 mol
step 2) Using the equation given, no. of mol of Cu required:
- 0.0179/2 X 3 = 0.026882 mol
step 3) since the solution given is in 1mol/L, the volume of CuSO4 required would be:
- 0.026882 X 1 = 0.026882L or 26.88ml
step 4) since u required 10% more, the volume would be:
-0.026882 X 110% = 0.02956989L or 29.56989mL.

That should be the answer u require tho i've gotta point out that i usually calculate the units in terms of and not mol/L  Bookmarks
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