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A sample of pynolusite (MnO_(2)) weights 0.5 gm. To this solutioin 0.594 gm As_(2)O_(3) and a dilute acid are added. After the reaction has stopped As^(+3) is AS_(2)O_(3) is titrated with 45 mlof M/50 KMn_(4) solution. Calculate the percentage of MnO_(2) in pyrolusite.

Answer» <html><body><p>`65.25%`<br/>`68%`<br/>`67.<a href="https://interviewquestions.tuteehub.com/tag/3-301577" style="font-weight:bold;" target="_blank" title="Click to know more about 3">3</a>%`<br/>`66.6%`</p>Solution :`MnO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>)+As_(2)O_(3)rarrMn^(+2)+AsO_(4)^(-3)` <br/> For excess of `As_(2)O_(3)`, we have <br/> `As_(2)O_(3)+MnO_(4)^(-)rarrMn^(+2)+AsO_(4)^(-3)` <br/> Npw `As_(2)O_(3)` getting oxidized to `AsO_(4)^(-3)`, the half reaction is <br/> `As_(2)O_(3)+5H_(2)Orarr2As+10H^(+)+4e^(-)` <br/> for <a href="https://interviewquestions.tuteehub.com/tag/1mol-283039" style="font-weight:bold;" target="_blank" title="Click to know more about 1MOL">1MOL</a> of `As_(2)O_(3)`, No. of `e^(-)s=4`, <br/> So, eq. wt of `Al_(2)O_(3)=198//4` <br/> M.eq of `As_(2)O_(3)=(0.594)/(198//4)xx1000=12` <br/> Me eq of excess of `As_(2)O_(3)` = M.eq of `KMnO_(4)` <br/> `=45=((1)/(50)xx5)=4.5` <br/> The other half reaction is <br/> `MnO_(4)^(-)+5e^(-)+8H^(+)rarrMn^(2)+4H_(2)O` <br/> M.eq of `As_(2)O_(3)` used for `MnO_(2)=12-4.5=7.5` <br/> `therefore` wt implies `(W)/(87//2)xx1000=7.5impliesW=0.326gm` <br/> `%MnO_(2)=(0.326)/(0.5)xx100=65.25%`</body></html>


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