Find the mass of potassium chlorate required | vedclass

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(C) The balanced chemical equation for the thermal decomposition of potassium chlorate is: $2KClO_3(s) \rightarrow 2KCl(s) + 3O_2(g)$ From the stoichi

Vedclass · Accepted Answer

$20.40$

Vedclass · Answer

(C) The balanced chemical equation for the thermal decomposition of potassium chlorate is: $2KClO_3(s) ightarrow 2KCl(s) + 3O_2(g)$ From the stoichiometry,$2 \ mol$ of $KClO_3$ produces $3 \ mol$ of $O_2$. At $STP$,$1 \ mol$ of any gas occupies $22.4 \ dm^3$. Therefore,$3 \ mol$ of $O_2$ occupies $3 	imes 22.4 \ dm^3 = 67.2 \ dm^3$. Mass of $2 \ mol$ of $KClO_3 = 2 	imes 122.5 \ g = 245 \ g$. Since $67.2 \ dm^3$ of $O_2$ is produced by $245 \ g$ of $KClO_3$,$5.6 \ dm^3$ of $O_2$ will be produced by: $\frac{245 \ g}{67.2 \ dm^3} 	imes 5.6 \ dm^3 = 20.416 \ g \approx 20.40 \ g$.

Find the mass of potassium chlorate required to liberate $5.6 \ dm^3$ of oxygen gas at $STP$. (Molar mass of $KClO_3 = 122.5 \ g/mol$) (in $g$)

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