Answer:
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!
