If you had 13.9 mL of a 0.0039 M solution of VO2+(aq), how many grams of Zn metal would be required to completely reduce the vanadium?

5,3x10⁻³g of Zn are required to reduce completely vanadium

Explanation:

The balanced redox reactions for the sequential reduction of vanadium are:

Reduction from +5 to +4:

2 VO₂⁺(aq) + 4 H⁺(aq) + Zn(s) → 2 VO²⁺(aq) + Zn²⁺(aq) + 2 H₂O(l)

reduction from +4 to +3:

2 VO²⁺(aq) + Zn(s) + 4 H⁺(aq) → 2 V³⁺(aq) + Zn²⁺(aq) + 2 H₂O(l)

reduction from +3 to +2:

2 V³⁺(aq) + Zn(s) → 2 V²⁺(aq) + Zn²⁺(aq)

The sum of the reactions to a complete reduction is:

2 VO₂⁺(aq) + 8H⁺(aq) + 3Zn(s) → 2 V²⁺(aq) + 3Zn²⁺(aq) + 4 H₂O(l)

13,9mL of a 0,0039M VO₂⁺ are:

0,0139L × (0,039mol/L) = 5,421x10⁻⁵ moles of VO₂⁺. As 2 moles of VO₂⁺ reacts with 3 moles of Zn. Moles of Zn that react are:

5,421x10⁻⁵ moles of VO₂⁺ × (3 moles Zn / 2 moles of VO₂⁺) = 8,1315x10⁻⁵ moles of Zn. As molar mass of Zn is 65,38g/mol:

8,1315x10⁻⁵ moles of Zn × (65,38g/mol) =

5,3x10⁻³g of Zn are required to reduce completely vanadium.

I hope it helps!