The decomposition of potassium chlorate ( $K C l O_{3}$ $KClO_3$ ) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?

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Answer 1 · 2017-02-25T08:24:06

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

Normally, this rxn must be calalyzed with a little $Mn(IV)$ salt, but we address the question with this stoichiometry.

We spit out a $19.7*mol$ quantity of dioxygen gas. Given the stoichiometry there were $2/3xx19.7*mol$ $"potassium chlorate"$ .

This represents a mass of $2/3xx19.7*molxx122.55*g*mol^-1$ of $"potassium chlorate"$ $~~$ $1.6*kg$ .

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The decomposition of potassium chlorate (KClO_3KClO3KClO_3) is used as a source of oxygen in the laboratory. How much potassium chlorate is needed to produce 19.7 mol of oxygen?