Predict the products of each of the following reactions and then balance the chemical equations.

(a) Fe is heated in an atmosphere of steam.

(b) NaOH is added to a solution of Fe(NO₃)₃.

(c) FeSO₄ is added to an acidic solution of KMnO₄.

(d) Fe is added to a dilute solution of H₂SO₄.

(e) A solution of Fe(NO₃)₂ and HNO₃ is allowed to stand in air.

(f) FeCO₃ is added to a solution of HClO₄.

(g) Fe is heated in air.

Redox is a type of chemical reaction in which the oxidation states of atoms are changed. Most often with one species undergoing oxidation while another species undergoes reduction.

The following equations are balanced in following manner:

(a)

\(Fe(s) + {H_2}O(g) \to F{e_3}{O_4}(s) + {H_2}(g)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:3Fe \to 3F{e^{\frac{8}{3} + }} + 8{e^ - }\\R:8{H^ + } + 8{e^ - } \to 8H\\\overline {3Fe + 8{H^ + } \to 3F{e^{\frac{8}{3} + }} + 8H} \end{aligned}\)

From, that we have,

\(3Fe + 4{H_2}O(g) \to F{e_3}{O_4}(s) + 4{H_2}(g)\)

(b)

\(NaOH(aq) + Fe{(N{O_3})_3}(aq) \to Fe{(OH)_3}(s) + N{a^ + }(aq) + NO_3^ - (aq)\)

After balancing the oxidation number and reduction number we have:

\(3NaOH(aq) + Fe{(N{O_3})_3}(aq) \to Fe{(OH)_3}(s) + N{a^ + }(aq) + 3NO_3^ - (aq)\)

From, that we have,

\(3NaOH(aq) + Fe{(N{O_3})_3}(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3NO_3^ - (aq)\)

(c)

\(F{e^{2 + }}(s) + MnO_4^ - (aq) + {H^ + }(aq) \to F{e^{3 + }}(aq) + M{n^{2 + }}(aq) + {H_2}O(l)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:5F{e^{2 + }} \to 5F{e^{3 + }} + 5{e^ - }\\R:MnO_4^ - + 5{e^ - } + 8{H^ + } \to M{n^{2 + }} + 4{H_2}O\\\overline {5F{e^{2 + }} + MnO_4^ - + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O(l)} \end{aligned}\)

From, that we have,

\(5F{e^{2 + }} + MnO_4^ - + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O(l)\)

(d)

\(Fe(s) + {H_3}{O^ + }(aq) + SO_4^{2 - }(aq) \to FeS{O_4}(aq) + {H_2}(g) + {H_2}O(l)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:Fe \to F{e^{2 + }} + 2{e^ - }\\R:2{H^ + } + 2{e^ - } \to 2{H^0}\\F{e^ + } + 2{H^ + } \to F{e^{2 + }} + 2{H^0}\end{aligned}\)

From, that we have,

\(Fe(s) + 2{H_3}{O^ + }(aq) + SO_4^{2 - }(aq) \to FeS{O_4}(aq) + {H_2}(g) + 2{H_2}O(l)\)

(e)

\(F{e^{2 + }}(aq) + {O_2}(g) + HN{O_3}(aq) \to F{e^{3 + }}(aq) + NO_3^ - (aq) + {H_2}O(l)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:4F{e^{2 + }} \to 4F{e^{3 + }} + 4{e^ - }\\R:2{O^0} + 4{e^ - } \to 2{O^{2 - }}\\4F{e^{2 + }} + 2{O^0} \to 4F{e^{3 + }} + 2{O^{2 - }}\end{aligned}\)

From, that we have,

\(4F{e^{2 + }}(aq) + {O_2}(g) + HN{O_3}(aq) \to 4F{e^{3 + }}(aq) + NO_3^ - (aq) + 2{H_2}O(l)\)

(f)

\(FeC{O_3}(s) + HCl{O_4}(aq) \to Fe{(Cl{O_4})_2}(aq) + C{O_2}(g) + {H_2}O(l)\)

From, that we have,

\(FeC{O_3}(s) + 2HCl{O_4}(aq) \to Fe{(Cl{O_4})_2}(aq) + C{O_2}(g) + {H_2}O(l)\)

(g)

\(Fe(s) + {O_2}(g) \to F{e_3}{O_4}(s)\)

After balancing the oxidation number and reduction number we have:

\(\begin{aligned}{l}O:3Fe \to 3F{e^{\frac{8}{3} + }} + 8{e^ - }\\R:2{O^0} + 4{e^ - } \to 2{O^{2 - }}\\\overline {3Fe + 4{O^0} \to 3F{e^{\frac{8}{3} + }} + 4{O^{2 - }}} \end{aligned}\)

From, that we have,

\(3Fe(s) + 2{O_2}(g) \to F{e_3}{O_4}(s)\)

Hence, the final answer is,

The balanced reactions are as follow:

a)\(3Fe + 4{H_2}O(g) \to F{e_3}{O_4}(s) + 4{H_2}(g)\)

b)\(3NaOH(aq) + Fe{(N{O_3})_3}(aq) \to Fe{(OH)_3}(s) + 3N{a^ + }(aq) + 3NO_3^ - (aq)\)

c)\(5F{e^{2 + }} + MnO_4^ - + 8{H^ + } \to 5F{e^{3 + }} + M{n^{2 + }} + 4{H_2}O(l)\)

d)\(Fe(s) + 2{H_3}{O^ + }(aq) + SO_4^{2 - }(aq) \to FeS{O_4}(aq) + {H_2}(g) + 2{H_2}O(l)\)

e)\(4F{e^{2 + }}(aq) + {O_2}(g) + HN{O_3}(aq) \to 4F{e^{3 + }}(aq) + NO_3^ - (aq) + 2{H_2}O(l)\)

f)\(FeC{O_3}(s) + 2HCl{O_4}(aq) \to Fe{(Cl{O_4})_2}(aq) + C{O_2}(g) + {H_2}O(l)\)

g) \(3Fe(s) + 2{O_2}(g) \to F{e_3}{O_4}(s)\)