A chemist performs the reaction #6ClO_2 + 3H_2O -> 5HClO_3 + HCl#. If a chemist wants to make 50.0 g of #HClO_3#, what is the minimum number of grams of #ClO_2# that she can use?

Question

2461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-21T14:15:23

Well, #"chlorine dioxide"# DISPROPORTIONATES....and I make it mass of UNDER #50*g# with respect to #ClO_2#.

#ClO_2# is oxidized to #HClO_3#:

#stackrel(IV)ClO_2(aq) +H_2O(l)rarr Hstackrel(V)ClO_3(aq)+H^+ +e^(-)# #(i)#

And #ClO_2# is reduced to #Cl^(-)#:

#stackrel(IV)ClO_2(aq)+4H^+ +5e^(-)rarr Cl^(-)+2H_2O(l)# #(ii)#

And for both half-equations, mass and charge are conserved. And so we take #5xx(i)+(ii)# to get....

#5ClO_2(aq)+ClO_2(aq)+4H^+ +5e^(-) +5H_2O(l)rarr 5HClO_3(aq)+Cl^(-)+2H_2O(l)+5H^+ +5e^(-)#

And we cancel common reagents....

#5ClO_2(aq)+ClO_2(aq) +3H_2O(l)rarr 5HClO_3(aq)+Cl^(-)+H^+ #

...or...

#6ClO_2(aq) +3H_2O(l)rarr HCl(aq)+5HClO_3(aq)#

The which I think is balanced with respect to mass and charge.

And so (finally!) we want a molar quantity with respect to #"chloric acid"#..of #(50.0*g)/(84.46*g*mol^-1)=0.592*mol#..

And so with respect to #ClO_2# we needs #6/5*"equiv"# ...

i.e. #6/5xx0.592*molxx67.45*g*mol^-1=??*g#