Answer:
First, write an equation for the reaction involved.
Fe + CuSO4 ---> FeSO4 + Cu
Now, calculate the no. of moles added in the reaction. (n.o.m. = mass /molar mass)
No. of moles of Fe added = 2 / 55.8
= 0.0358 mol
No. of moles of CuSO4 = 7 / (63.5 +32.1+16x4)
= 0.04386 mol
From the equation, the mole ratio of Fe : CuSO4 = 1:1,
meaning 1 mole of Fe reacts with 1 mole of CuSO4.
It also indicates that the no. of moles reacted in Fe equals to the no. of moles reacted in CuSO4.
0.04386 > 0.0358
The no. of moles of CuSO4 is higher than that of Fe, meaning all moles of Fe will be reacted while not all CuSO4 reacts.
Hence Fe is the limiting reagent, and CuSO4 is in excess.
All 0.0358 moles of Fe reacts, meaning the no. of moles of CuSO4 left unreacted = 0.04386 - 0.0358
= 0.0080596mol
mass of CuSO4 unreacted = 0.0080596 x (63.5 +32.1+16x4)
= 1.28g
