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?

Apr 21, 2018

Well, $\text{chlorine dioxide}$ DISPROPORTIONATES....and I make it mass of UNDER $50 \cdot g$ with respect to $C l {O}_{2}$.

Explanation:

$C l {O}_{2}$ is oxidized to $H C l {O}_{3}$:

$\stackrel{I V}{C} l {O}_{2} \left(a q\right) + {H}_{2} O \left(l\right) \rightarrow H \stackrel{V}{C} l {O}_{3} \left(a q\right) + {H}^{+} + {e}^{-}$ $\left(i\right)$

And $C l {O}_{2}$ is reduced to $C {l}^{-}$:

$\stackrel{I V}{C} l {O}_{2} \left(a q\right) + 4 {H}^{+} + 5 {e}^{-} \rightarrow C {l}^{-} + 2 {H}_{2} O \left(l\right)$ $\left(i i\right)$

And for both half-equations, mass and charge are conserved. And so we take $5 \times \left(i\right) + \left(i i\right)$ to get....

$5 C l {O}_{2} \left(a q\right) + C l {O}_{2} \left(a q\right) + 4 {H}^{+} + 5 {e}^{-} + 5 {H}_{2} O \left(l\right) \rightarrow 5 H C l {O}_{3} \left(a q\right) + C {l}^{-} + 2 {H}_{2} O \left(l\right) + 5 {H}^{+} + 5 {e}^{-}$

And we cancel common reagents....

$5 C l {O}_{2} \left(a q\right) + C l {O}_{2} \left(a q\right) + 3 {H}_{2} O \left(l\right) \rightarrow 5 H C l {O}_{3} \left(a q\right) + C {l}^{-} + {H}^{+}$

...or...

$6 C l {O}_{2} \left(a q\right) + 3 {H}_{2} O \left(l\right) \rightarrow H C l \left(a q\right) + 5 H C l {O}_{3} \left(a q\right)$

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

And so (finally!) we want a molar quantity with respect to $\text{chloric acid}$..of $\frac{50.0 \cdot g}{84.46 \cdot g \cdot m o {l}^{-} 1} = 0.592 \cdot m o l$..

And so with respect to $C l {O}_{2}$ we needs $\frac{6}{5} \cdot \text{equiv}$ ...

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