Question

Could someone help me to solve the problem?

In a beaker `2.5g` of `"Fe"` and dilute `"H"_2"SO"_4` is taken. The volume of this solution is `500ml`. In another beaker `30ml` of `0.015"M"` `"KMnO"_4` is taken. we mix `50ml` of the first solution.

(1) If `"HCl"` was used instead of `"H"_2"SO"_4`, could we measure the amount of iron accurately?

(2) Would the volume of `"K"_2"Cr"_2"O"_7` be equal to the volume of `"KMnO"_4`, if we used `"K"_2"Cr"_2"O"_7` instead of `"KMnO"_7`?

Answer 1

(1) Yes, you could still measure the amount of iron accurately.
(2) No, the volume of `"K"_2"Cr"_2"O"_7` would not be equal to the volume of `"KMnO"_4`,

Explanation:

(1) `"HCl"` vs. `"H"_2"SO"_4`

It doesn't matter if you use `"HCl"` or `"H"_2"SO"_4`.

Their only function is to provide the hydrogen ions to react with the iron.

`"Fe(s) + 2H"^"+""(aq)" → "Fe"^"2+""(aq)" + "H"_2"(g)"`

Both `"HCl"` and `"H"_2"SO"_4` are strong acids, so they will each work equally well.

(2) `"KMnO"_4` vs. `"K"_2"Cr"_2"O"_7`

`"MnO"_4^"-" + 8"H"^"+" + 7"e"^"-" → "Mn"^"2+" + 4"H"_2"O"`

`"Cr"_2"O"_7^"2-" + 14"H"^"+" + 6"e"^"-" → "2Cr"^"3+" + 7"H"_2"O"`

1 mol of `"KMnO"_4` will accept 7 mol of electrons, while
1 mol of `"K"_2"Cr"_2"O"_7` will accept only 6 mol of electrons.

If you are using equal molar concentrations of each, the volume of `"K"_2"Cr"_2"O"_7` needed to oxidize a given amount of `"Fe"^"2+"` will be greater than the volume of `"KMnO"_4`.