When the following equation is balanced in an acidic solution how many moles of water are present?

Reduction half equation: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l) …… [1]
Oxidation half equation: Fe²⁺(aq) → Fe³⁺(aq) + e⁻ …… [2]

[1] + [2]×6, and cancel 6e⁻ on the both sides. The balanced equation is :
Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 6Fe³⁺(aq) + 7H₂O(l)

Number of moles of water (H₂O) present = 7 mol

1. Divide reaction in to 2 half-reactions and balance metal atom:
Fe2+ --> Fe3+
Cr2O72- --> 2 Cr3+

2. Balance O by adding H2O:
Fe2+ --> Fe3+
Cr2O72- --> 2 Cr3+ + 7 H2O

3. Balance H by adding H+:
Fe2+ --> Fe3+
Cr2O72- + 14 H+ --> 2 Cr3+ + 7 H2O

4. Balance charges in each half-reaction by adding electrons (e-):
Fe2+ --> Fe3+ + e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O

5. Multiply one or both half-reactions to balance electrons:
6 Fe2+ --> 6 Fe3+ + 6 e-
Cr2O72- + 14 H+ + 6 e- --> 2 Cr3+ + 7 H2O

6. Add the two half-reactions and simplify as needed to give balanced equation:
6 Fe2+ + Cr2O72- + 14 H+ --> 6 Fe3+ + 2 Cr3+ + 7 H2O

The answer is 7.