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Question: Calculate the mass \[\left( {\mathbf{m}} \right)\] of oxalic acid \[\left( {{{\mathbf{H}}_2}{{\mathb...

Calculate the mass (m)\left( {\mathbf{m}} \right) of oxalic acid (H2C2O4)\left( {{{\mathbf{H}}_2}{{\mathbf{C}}_2}{{\mathbf{O}}_4}} \right) which can be oxidised to CO2{\mathbf{C}}{{\mathbf{O}}_2}by 100mL{\mathbf{100mL}} of MnO4{\mathbf{Mn}}{{\mathbf{O}}_4}^ - solution, 10mL{\mathbf{10mL}} of which is capable of oxidising 50.0mL{\mathbf{50}}.{\mathbf{0mL}} of 1.00N{\mathbf{1}}.{\mathbf{00N}} I{I^ - } to I2{I_2}.
MnO4+8H++5eMn2++4H2O{\mathbf{Mn}}{{\mathbf{O}}_4}^ - + {\mathbf{8}}{{\mathbf{H}}^ + } + {\mathbf{5}}{{\mathbf{e}}^ - } \to {\mathbf{M}}{{\mathbf{n}}_2}^ + + {\mathbf{4}}{{\mathbf{H}}_2}O
H2C2O42CO2+2H++2e{{\mathbf{H}}_2}{{\mathbf{C}}_2}{{\mathbf{O}}_4} \to {\mathbf{2C}}{{\mathbf{O}}_2} + {\mathbf{2}}{{\mathbf{H}}^ + } + {\mathbf{2}}{{\mathbf{e}}^ - }
2II2+2e{\mathbf{2}}{{\mathbf{I}}^ - } \to {{\mathbf{I}}_2} + {\mathbf{2}}{{\mathbf{e}}^ - }
What is the value of 10m?

Explanation

Solution

This is a pretty tough question to answer. This includes the stereochemistry of redox reaction. We need to consider two-equations, the first one is the oxidation of oxalic acid using potassium permanganate to form carbon dioxide and the second one is the oxidation of iodide ion to Iodine gas using potassium permanganate.

Complete step by step answer:
We are provided with (m)\left( m \right) mass of oxalic acid which is oxidized to carbon dioxide by the help of potassium permanganate which is 100mL100mL.
Now in the other sentence, we are given that 10mL10mL of this potassium permanganate is capable of oxidizing 50mL50mL of 1N1N solution iodide ion to iodine gas.
The normality of potassium permanganate solution in both the equation is the same, using the formula to find the unknown normality of solution. We can find the normality of potassium permanganate.
N1V1=N2V2(1){N_1}{V_1} = {N_2}{V_2} - \left( 1 \right)
N1{N_1} is normality of MnO4Mn{O_4}^ - (unknown)
V1{V_1}is volume of MnO4Mn{O_4}^ - (10mL)\left( {10mL} \right)
N2{N_2} is normality of iodine (1N)\left( {1N} \right)
V2{V_2} is volume of iodine (50mL)\left( {50mL} \right)
Putting the values in the above equation we get,
N110=150{N_1}*10 = 1*50
\Rightarrow N1=50/10{N_1} = 50/10
\Rightarrow N1=5N{N_1} = 5N
The normality of potassium permanganate is 5N5N.
Now we will calculate the number of equivalents of potassium permanganate, which can be calculated as
Normalitysolution(mL)/1000Normality*solution\left( {mL} \right)/1000
For potassium permanganate we get,
=5100/1000= 5*100/1000
=5/10= 5/10
=0.5= 0.5
Therefore the number of equivalents of oxalic acid is also 0.50.5.
Now to find the mass of oxalic acid we need to multiply its number of equivalents to its molecular mass,
The molecular mass of oxalic acid is 45g45g
\Rightarrow m=0.545m = 0.5*45
\Rightarrow m=22.5gm = 22.5g
In the question we are asked for 10m10m, so
10m=1022.510m = 10*22.5
\Rightarrow 10m=225g10m = 225g

Note:
In the question, we need to understand how one molecule is important in one reaction and the same molecule is important in the other reaction. If it is the same molecule which reacts in both the reactions, we can find the similarities between both the reaction and then proceed to get to the answer.