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?

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

A chemist performs the reaction 6ClO_2 + 3H_2O -> 5HClO_3 + HCl6ClO_2 + 3H_2O -> 5HClO_3 + HCl. If a chemist wants to make 50.0 g of HClO_3HClO_3, what is the minimum number of grams of ClO_2ClO_2 that she can use?