The balanced redox reactions for the sequential reduction of vanadium are given below. 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) if you had 11.7 ml of a 0.0037 m solution of vo2+(aq), how many grams of zn metal would be required to completely reduce the vanadium

The overall equation for the combined reactions become:
2VO₂⁺²₍aq₎ + 8H⁺₍aq₎ + 3Zn⁺²⁽s⁾ ⇒ 2V⁺²₍aq₎ + 3Zn⁺²₍aq₎ + 4H₂O₍l₎
The volume of solution is:
11.7/1000 = 0.0117 Litres
The moles of VO₂⁺² are:
0.0117 × 0.0037 = 4.3 × 10⁻⁵ mol
As per the equation, 2 moles of VO₂⁺² need 3 moles of 3Zn⁺²
Therefore, moles of Zn⁺² needed are:
6.5 × 10⁻⁵ mol
One mole Zn metal produces one mol of ions. So we need
6.5 × 10⁻⁵ mol of Zn metal
Mass required = moles × Molecular weight
Mass = 6.5 × 10⁻⁵ × 65
Mass = 0.0042 grams

7is b8 is d9 is c10 is a11 is d

257 grams h2o i think