Answer:
[tex]M=0.120M[/tex]
Explanation:
Hello,
In this case, the undergone chemical reaction is:
[tex]MnO_4^-(aq)+H_2C_2O_4(aq)\rightarrow Mn^{+2}+CO_2[/tex]
In such a way, the acidic redox balance turns out:
[tex](Mn^{+7}O_4)^-+5e^-+8H^+\rightarrow Mn^{+2}+4H_2O\\H_2C_2O_4\rightarrow2CO_2+2H^++2e^-[/tex]
Which leads to the total balanced equation as follows:
[tex]2(MnO_4)^-(aq)+6H^+(aq)+5H_2C_2O_4(aq)\rightarrow2Mn^{+2}(aq)+8H_2O(l)+10CO_2(g)[/tex]
Thus, as the mass of oxalic acid is not given, one could suppose a value of 1 g (which you can modify based on the actual statement) in order to compute the oxalic acid moles as shwon below:
[tex]1gH_2C_2O_4*\frac{1molH_2C_2O_4}{90.04gH_2C_2O_4} *\frac{2mol(MnO_4)^-}{5molH_2C_2O_4} =0.00444mol(MnO_4)^-[/tex]
Whereby the molality results:
[tex]M=\frac{0.00444mol(MnO_4)^-}{0.03702L} =0.120M[/tex]
Remember you can modify the oxalic acid mass as you desire.
Best regards.
