Someone please help?

iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.547 grams of iron and 10.5 mL of 0.526M copper(II) sulfate react to form as much product as possible, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction?

Question

Someone please help?

 iron is oxidized to Fe2+ by a copper(II) sulfate solution, and 0.547 grams of iron and 10.5 mL of 0.526M copper(II) sulfate react to form as much product as possible, how many millimoles (mmol) of the non-limiting reactant will remain unused at the end of the reaction?

anonymous · Answer

Fe + CuSO4 = Cu + FeSO4

anonymous · Answer

molar mass of Fe is 55.85. 
0.547 g of Fe.
0.0098 mol of Fe

anonymous · Answer

10.5 mL = 0.0105 L
0.0105 x 0.526 = 0.0055 mol of CuSO4

anonymous · Answer

0.0055 < 0.0098 
Therefore CuSO4 is the limiting reagent

anonymous · Answer

0.0098 - 0.0055 = 0.004294

anonymous · Answer

4.29 mmol of CuSO4 remains unused in the reaction!

anonymous · Answer

Sorry I just looked at this now, I figured it out haha. I wasn't realizing that the volume was in mL.

anonymous · Answer

but thankyou

anonymous · Answer

no problem