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Step 1: Write the Balanced Thermochemical Reaction
When sodium chlorate ($NaClO_3$) is heated, it decomposes to sodium chloride ($NaCl$) and oxygen gas ($O_2$). The balanced reaction is:
$2\,NaClO_3 \,\xrightarrow{\Delta}\,2\,NaCl \,+\,3\,O_2$
Step 2: Determine the Moles of $O_2$ Released
It is given that $0.16\,g$ of oxygen ($O_2$) is released. The molar mass of $O_2$ is $32\,g\,mol^{-1}$. Hence,
$\text{Moles of }O_2 = \dfrac{0.16}{32} = 5\times 10^{-3}\,\text{moles}$
Step 3: Use Stoichiometric Ratios to Find Moles of $NaCl$ Produced
From the balanced equation, for every $3$ moles of $O_2$ produced, $2$ moles of $NaCl$ are formed. Thus,
$\dfrac{n(NaCl)}{2} \;=\;\dfrac{n(O_2)}{3}$
Substitute $n(O_2)=5\times 10^{-3}$:
$n(NaCl) = \dfrac{2}{3}\times 5\times 10^{-3} = \dfrac{10}{3}\times 10^{-3} = \dfrac{1}{300}\,\text{moles}$
Step 4: React $NaCl$ with $Ag^+$ to Form $AgCl$
The residue (sodium chloride) is then dissolved in water and treated with silver ions ($Ag^+$), forming silver chloride ($AgCl$):
$NaCl \;+\; Ag^+ \;\to\; AgCl \;+\; Na^+$
Each mole of $NaCl$ will produce one mole of $AgCl$. Therefore, the moles of $AgCl$ formed are equal to the moles of $NaCl$:
$n(AgCl) = n(NaCl) = \dfrac{1}{300}\,\text{moles}$
Step 5: Calculate the Mass of $AgCl$ Formed
The molar mass of $AgCl$ is given as $143.5\,g\,mol^{-1}$. Hence, the mass of $AgCl$ is:
$\text{Mass of }AgCl = 143.5 \times \dfrac{1}{300} \;=\; 0.48\,g$
Final Answer
$\boxed{0.48\,\text{g}}$ of $AgCl$ is formed.
