Question

Consider the unbalanced redox reaction: MnO?4(aq)+Zn(s)?Mn2+(aq)+Zn2+(aq) Balaced: 2MnO4?(aq)+16H+(aq)+5Zn(s)?2Mn2+(aq)+8H2O(l)+5Zn2+(aq) *Determine the volume of a 0.46M KMnO4 solution required to completely react with 2.8g of Zn.

Answer 1

The redox reaction is as folllows

2[MnO₄] ^- (aq) + 5Zn (s) + 16H⁺   2Mn²⁺ (aq) + 5Zn ²⁺ (aq) + 8H₂O (l)

Therefore 2 moles of KMnO4 will react with 5 moles of Zn

Now MW of Zn is 65.38

So, 2.8 g of Zn = 0.0428 mole

Consequently, 0.0428 mole of Zn will react with (2 x 0.0428) /5 = 0.0171 mole of KMnO4

0.46 M KMnO₄ means, 0.46 mole of  KMnO₄ is present in 1000 mL

Then, 0.0171 mole of KMnO4 is present in ( 1000 x 0.0171 ) /0.46 = 37.2 mL

So, 37.2 mL of 0.46M of KMnO4 is required to completely react with 2.8 g of Zn (ANSWER)