Question

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?

Answer 1

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

Explanation:

`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`