How many moles of potassium chlorate (KClO_3) | vedclass

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(B) The balanced chemical equation for the decomposition of potassium chlorate is: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$ At $NTP$,$1 \ mole$ of a

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$1/3$

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(B) The balanced chemical equation for the decomposition of potassium chlorate is: $2KClO_3(s) ightarrow 2KCl(s) + 3O_2(g)$ At $NTP$,$1 \ mole$ of any gas occupies $22.4 \ L$. Therefore,the number of moles of $O_2$ produced is: $n(O_2) = \frac{11.2 \ L}{22.4 \ L/mol} = 0.5 \ mol$. From the stoichiometry of the reaction,$3 \ moles$ of $O_2$ are produced from $2 \ moles$ of $KClO_3$. So,$1 \ mole$ of $O_2$ is produced from $\frac{2}{3} \ moles$ of $KClO_3$. Thus,$0.5 \ moles$ of $O_2$ will be produced from: $n(KClO_3) = \frac{2}{3} 	imes 0.5 = \frac{2}{3} 	imes \frac{1}{2} = \frac{1}{3} \ mol$.

How many moles of potassium chlorate $(KClO_3)$ must be heated to obtain $11.2 \ L$ of oxygen at $NTP$?

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