How many moles of potassium chlorate (KClO_3) | vedclass

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(B) Step $1$: Write the balanced chemical equation for the thermal decomposition of potassium chlorate:$2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$Step

Vedclass · Accepted Answer

$1/3 	ext{ mole}$

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(B) Step $1$: Write the balanced chemical equation for the thermal decomposition of potassium chlorate:$2KClO_3(s) ightarrow 2KCl(s) + 3O_2(g)$Step $2$: From the stoichiometry, $2 	ext{ moles}$ of $KClO_3$ produce $3 	ext{ moles}$ of $O_2$.Step $3$: At $STP$, $1 	ext{ mole}$ of any gas occupies $22.4 	ext{ L}$. Therefore, $11.2 	ext{ L}$ of $O_2$ corresponds to $n = \frac{11.2 	ext{ L}}{22.4 	ext{ L/mol}} = 0.5 	ext{ moles}$ of $O_2$.Step $4$: Using the mole ratio, $3 	ext{ moles}$ of $O_2$ are produced by $2 	ext{ moles}$ of $KClO_3$. So, $0.5 	ext{ moles}$ of $O_2$ are produced by $\frac{2}{3} 	imes 0.5 = \frac{2}{3} 	imes \frac{1}{2} = \frac{1}{3} 	ext{ moles}$ of $KClO_3$.

How many moles of potassium chlorate $(KClO_3)$ are required to be heated to produce $11.2 \text{ L}$ of oxygen gas at $STP$?

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