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?

1 Answer
Ernest Z.
Nov 27, 2017

(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.

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