Solveeit Logo

Question

Question: A sample of hydrazine sulphate (\[{N_2}{H_6}S{O_4}\]​) was dissolved in \[100\;mL\;\] of water. \[10...

A sample of hydrazine sulphate (N2H6SO4{N_2}{H_6}S{O_4}​) was dissolved in 100  mL  100\;mL\; of water. 10  mL  10\;mL\; of this solution was reacted with an excess of FeCl3FeC{l_3} ​ solution and warmed to complete the reaction. Ferrous ion formed was estimated and it required 20  mL  20\;mL\; of M50KMnO4\dfrac{M}{{50}}KMn{O_4} ​. Estimate the amount of hydrazine sulphate in one litre of solution.
Reactions: 4Fe3++N2H4N2+4Fe2++4H+4F{e^{3 + }} + {N_2}{H_4}\, \to {N_2}\, + \,\,4F{e^{2 + }} + \,\,4{H^ + }
MnO4+5Fe2++8HMn2++5Fe3++4H2O.MnO_4^ - + \,\,5F{e^{2 + }} + \,\,8{H^ - } \to M{n^{2 + }} + \,\,5F{e^{3 + }} + \,\,4{H_2}O.

Explanation

Solution

To calculate the amount of Hydrazine Sulphate that’s gets dissolved in one litre of solution, we will compare the Equivalent mass of hydrazine Sulphate (N2H6SO4{N_2}{H_6}S{O_4}) with equivalent mass of potassium permanganate (KMnO4KMn{O_4} ).

Complete step by step answer:
Given redox reactions:
4Fe3++N2H4N2+4Fe2++4H+4F{e^{3 + }} + {N_2}{H_4}\, \to {N_2}\, + \,\,4F{e^{2 + }} + \,\,4{H^ + }
MnO4+5Fe2++8HMn2++5Fe3++4H2O.MnO_4^ - + \,\,5F{e^{2 + }} + \,\,8{H^ - } \to M{n^{2 + }} + \,\,5F{e^{3 + }} + \,\,4{H_2}O.
The redox changes are as follows:
For FeCl3FeC{l_3}: Fe3+Fe2++1eF{e^{3 + }} \to \,F{e^{2 + }} + 1e
For N2H6SO4{N_2}{H_6}S{O_4}: N22N2+4eN_2^{2 - } \to {N_2} + 4e
Now, we will compare the equivalent mass of Potassium permanganate and equivalent mass of Hydrazine Sulphate in 10  mL  10\;mL\; solution:
MeqN2H6SO4=MeqKMnO4{M_{eq}}\,{N_2}{H_6}S{O_4} = {M_{eq}}\,KMn{O_4}
The molar mass of hydrazine sulphate is 130g130\,g and the change in oxidation is 4 so its n factor will be 44.
As the n factor is 44the equivalent Mass of Hydrazine Sulphate will be=1304=32.5 = \dfrac{{130}}{4} = 32.5
Now, we will calculate the amount of hydrazine present in 10  mL  10\;mL\; solution. So the amount will be;
Amountofhydrazine=110×32.51000×20=0.065gAmount\,of\,hydrazine = \dfrac{1}{{10}} \times \dfrac{{32.5}}{{1000}} \times 20 = 0.065\,g
Now, with this, we will calculate the amount of hydrazine sulphate present in one litre of solution. It will be;
Amountofhydrazinein1lsolution=0.6510×1000=6.5gAmount\,of\,hydrazine\,in\,1l\,solution = \dfrac{{0.65}}{{10}} \times 1000 = 6.5\,g
Hence, the amount of hydrazine sulphate required in one litre of solution will be 6.5g6.5\,g.

Additional information
Hydrazine sulphate is a salt of the hydrazinium cation and the anion of bi-sulfate. It is a water-soluble white salt. It is used as a catalyst in fiber making from acetate. It is also used industrially as jet fuel and in the treatment of cancer. N2H6SO4{N_2}{H_6}S{O_4} can also be used as a fungicide and antiseptic. You can also prepare N2H6SO4{N_2}{H_6}S{O_4} by reacting hydrazine with sulphuric acid.

Note:
KMnO4KMn{O_4} or Potassium permanganate is an organic compound. It is used as an oxidizing agent. It is a purple-black crystalline solid in colour. It reduces itself and oxidizes the other compounds. It is soluble in water and is also used for cleaning of wounds.