How many grams of $KCIO_3$ must be heated to produce 6.8 moles of oxygen?

Question

1434 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-04T16:33:27

Over $500*g$ $KClO_3(s)$ are required; I assume you mean $"potassium chlorate"$ .

We assess the decomposition reaction of $"potassium chlorate"$ :

$KClO_3(s) +Deltararr 3/2O_2(g) + 2KCl(s)$

This reaction is typically catalyzed by $MnO_2$ , which acts as an oxygen transfer agent.

And thus each equiv dioxygen gas requires $2/3$ equiv of $"potassium chlorate"$ . Per equiv $"potassium chlorate"$ , how much dioxygen do we get?

We require $6.8*mol$ $O_2(g)$ and thus we need $2/3xx6.8*mol$ $"potassium chlorate"$ .

i.e. $2/3xx6.8*molxx122.55*g*mol^-1~=0.6*kg$

How many grams of KCIO_3KCIO_3 must be heated to produce 6.8 moles of oxygen?