A given initial mass of KClO_3 on 50% decompo | vedclass

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(C) The balanced chemical equation for the decomposition of $KClO_3$ is: $2KClO_3 \rightarrow 2KCl + 3O_2$ From the stoichiometry,$2 \ mol$ of $KClO_3

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$490$

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(C) The balanced chemical equation for the decomposition of $KClO_3$ is: $2KClO_3 ightarrow 2KCl + 3O_2$ From the stoichiometry,$2 \ mol$ of $KClO_3$ produces $3 \ mol$ of $O_2$. Given volume of $O_2 = 67.2 \ L$ at $STP$ $(0 \ ^\circ C, 1 \ atm)$. Since $1 \ mol$ of gas occupies $22.4 \ L$ at $STP$,Moles of $O_2 = \frac{67.2 \ L}{22.4 \ L/mol} = 3 \ mol$. Using the stoichiometry,$3 \ mol$ of $O_2$ is produced by $2 \ mol$ of $KClO_3$. So,$2 \ mol$ of $KClO_3$ decomposed. Given that only $50\%$ of the initial mass of $KClO_3$ decomposed,let the initial mass be $m \ g$. Moles of $KClO_3$ decomposed = $\frac{0.5 	imes m}{122.5 \ g/mol} = 2 \ mol$. $0.5 	imes m = 2 	imes 122.5 = 245$. $m = \frac{245}{0.5} = 490 \ g$.

$A$ given initial mass of $KClO_3$ on $50\%$ decomposition produces $67.2 \ L$ of oxygen gas at $0 \ ^\circ C$ and $1 \ atm$. The other product of decomposition is $KCl$. The initial mass of $KClO_3$ (in $g$) taken is:

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