A sample of NaClO_3 is converted by heat to N | vedclass

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(D) The balanced chemical equation for the thermal decomposition of $NaClO_3$ is:$2NaClO_3 \xrightarrow{\Delta} 2NaCl + 3O_2$Number of moles of $O_2$

Vedclass · Accepted Answer

$0.48$

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(D) The balanced chemical equation for the thermal decomposition of $NaClO_3$ is:$2NaClO_3 \xrightarrow{\Delta} 2NaCl + 3O_2$Number of moles of $O_2$ produced from $0.16 \ g$ of $O_2$ (molar mass = $32 \ g \ mol^{-1}$):$n(O_2) = \frac{0.16 \ g}{32 \ g \ mol^{-1}} = 0.005 \ mol$From the stoichiometry of the reaction,$3 \ mol$ of $O_2$ is produced along with $2 \ mol$ of $NaCl$.Therefore,$0.005 \ mol$ of $O_2$ corresponds to:$n(NaCl) = n(O_2) 	imes \frac{2}{3} = 0.005 	imes \frac{2}{3} = 0.00333 \ mol$Since $NaCl$ reacts with $AgNO_3$ to form $AgCl$ in a $1:1$ molar ratio $(NaCl + AgNO_3 ightarrow AgCl + NaNO_3)$:$n(AgCl) = n(NaCl) = 0.00333 \ mol$Mass of $AgCl$ obtained:$Mass = n(AgCl) 	imes 	ext{Molar mass}(AgCl) = 0.00333 \ mol 	imes 143.5 \ g \ mol^{-1} \approx 0.478 \ g \approx 0.48 \ g$

$A$ sample of $NaClO_3$ is converted by heat to $NaCl$ with a loss of $0.16 \ g$ of oxygen. The residue is dissolved in water and precipitated as $AgCl$. The mass of $AgCl$ (in $g$) obtained will be (Given: Molar mass of $AgCl = 143.5 \ g \ mol^{-1}$)

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