Question

Basic solutions of \({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}}\) are powerful oxidants . What mass of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{0}}\) reacts with 125.0 ml of 0.1717 M basic solution of \({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}}\) that contains an excess of Sodium Hydroxide if the products include Xe and the solution of Sodium Permanganate ?

Short answer

34.4 milimoles of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\) .

Step-by-step solution

Calculating valence :

The oxidation state (o.s) of Mn in \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}{\rm{.6}}{{\rm{H}}_{\rm{2}}}{\rm{0}}\) is = +2

[ Let o.s of Mn is x ,\(x + 2( - 1) = 0\)or,\(x = + 2\)]

\(\)\({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}} \to {\rm{4N}}{{\rm{a}}^{\rm{ + }}}{\rm{ + Xe0}}_{\rm{6}}^{{\rm{4 - }}}\); \({\rm{4N}}{{\rm{a}}^{\rm{ + }}}\) = 4 ×(+1) = + 4 , 1\({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\)= 1×(-4) = - 4 \(\)

∴ Valence = (4+4) = 8 [ Neglecting + and – sign ]

Finding the amount of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\) :

According to question we can write, \({\rm{M}}{{\rm{n}}^{{\rm{2 + }}}}{\rm{ + Xe0}}_{\rm{6}}^{{\rm{4 - }}} \to {\rm{Xe + Mn0}}_{\rm{4}}^{\rm{ - }}\)

Milimoles of \({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\) = \(125 \times 0.1717 = 21.5\)

Miliequivalents of \({\rm{N}}{{\rm{a}}_{\rm{4}}}{\rm{Xe}}{{\rm{0}}_{\rm{6}}}\) = \(\) valence × milimoles

=\(8 \times 21.5 = 172\)meq

Miliequivalents of = meq. Of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\) = 172 meq

Milimoles = \(\) \(\frac{{{\rm{miliequivalents}}}}{{{\rm{n - factor}}}}\)

[\({\rm{4N}}{{\rm{a}}^{\rm{ + }}}\)and \({\rm{XeO}}_{\rm{6}}^{{\rm{4 - }}}\) i.e. n factor = ( 4 +1 ) = 5 ]

Milimoles of \({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\) = \(\frac{{{\rm{172}}}}{{\rm{5}}}{\rm{ = 34}}{\rm{.4}}\) milimoles

Hence , the required mass of\({\rm{Mn(N}}{{\rm{0}}_{\rm{3}}}{{\rm{)}}_{\rm{2}}}\)is 34.4 milimoles.