Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A sample of $MnSO_44H_2O$ is heated in air to give $Mn_3O_4$.The residue $Mn_3O_4$ is dissolved in 100ml of $N/10$ $FeSO_4$ containing $H_2SO_4$.The solution reacts completely with 50ml of $KMnO_4$.25ml of this $KMnO_4$ requires 30ml of N/10 $FeSO_4$ for complete oxidation.Determine the amount of $MnSO_4.4H_2O$ is the sample


Can you answer this question?

1 Answer

0 votes
$MnSO_4.4H_2O\qquad\underrightarrow{\Delta}\qquad Mn_3O_4$+dissolved in 100ml of N/10 $FeSO_4$
Meq of $FeSO_4$ taken=$\large\frac{1}{10}$$\times 100=10$
The solution further requires 50ml of $KMnO_4$,an oxidant (of unknown N),hence $FeSO_4$ is in excess
Excess of Meq of $FeSO_4$=Meq of $KMnO_4$(50ml)
25ml of N-$KMnO_4$=30ml of N/10 $FeSO_4$
$25\times N=30\times \large\frac{1}{10}$$\Rightarrow N=\large\frac{3}{25}$
Excess meq of $FeSO_4=\large\frac{3}{25}$$\times 50=6$
Meq of $FeSO_4$ used against $Mn_3O_4$=10-6=4
Also,meq of $Mn_3O_4$=Meq of $MnSO_4$
$\large\frac{g}{E_{MnSO_4}}$$\times 1000=4$
$\Rightarrow g=\large\frac{4}{1000}$$\times E_{MnSO_4}$
$3Mn^{2+}+4H_2O\rightarrow Mn_3O_4+8H6++2e^-$
$x=\large\frac{2}{3}\Rightarrow E=\large\frac{223}{\Large\frac{2}{3}}$
$\Rightarrow g=\large\frac{4}{1000}\times \frac{223}{\Large\frac{2}{3}}$
$\Rightarrow 1.338gm$
Hence (c) is the correct answer.
answered Jan 8, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App