An iron nail weighing 40.0 g is placed in 200.0 g of copper(II) sulfate, which reacts to form pure copper and iron(III) sulfate by the following reaction: __Fe(s) + __CuSO4(s) → __Fe2(SO4)3(s) + __Cu(s)
a. Fill in the missing coefficients to balance the equation.
b. What is the limiting reagent and the mass of iron(III) sulfate that is formed?
c. After the reaction goes to completion, what is the mass of the reactant that remains?

Fe(s) + CuSO4 (aq) ! FeSO4 (aq) + Cu(s)

Fe(s) + Cu+2(aq) ! Fe+2(aq) + Cu(s)

(2) 2Fe(s) + 3CuSO4 (aq) ! Fe2(SO4)3 (aq) + 3Cu(s)

2Fe(s) + 3Cu+2(aq) ! 2Fe+3(aq) + 3Cu(s)

Explanation:

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explanation: