In the reaction 2 KClO_{3(s)} longrightarrow | vedclass

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(D) The balanced chemical equation is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$From the stoichiometry of the reaction,$3 \ moles$ of $

Vedclass · Accepted Answer

$74.5$

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(D) The balanced chemical equation is: $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$From the stoichiometry of the reaction,$3 \ moles$ of $O_2$ are produced along with $2 \ moles$ of $KCl$.At $S.T.P.$,$1 \ mole$ of any gas occupies $22.4 \ L$.Thus,$3 \ moles$ of $O_2$ occupy $3 	imes 22.4 \ L = 67.2 \ L$.The molar mass of $KCl = 39 + 35.5 = 74.5 \ g \ mol^{-1}$.So,$2 \ moles$ of $KCl = 2 	imes 74.5 \ g = 149 \ g$.According to the reaction,$67.2 \ L$ of $O_2$ is produced by $149 \ g$ of $KCl$.Therefore,$33.6 \ L$ of $O_2$ will be produced by: $\frac{149 	imes 33.6}{67.2} = 74.5 \ g$ of $KCl$.

In the reaction $2 KClO_{3(s)} \longrightarrow 2 KCl_{(s)} + 3 O_{2(g)}$,$\Delta H^{\circ} = -78 \ kJ$. If $33.6 \ L$ of oxygen gas is liberated at $S.T.P.$,what is the mass of $KCl_{(s)}$ produced (in $g$)? (Atomic mass: $K = 39, Cl = 35.5 \ g \ mol^{-1}$)

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