91Ó°ÊÓ

Potassium chlorate decomposes to produce potassium chloride and oxygen. The balanced equation is like

$2KCI{O}_3\left(s\right)\stackrel{∆}{→}2KCI\left(s\right)+3O_2\left(g\right)$

As you know,

Mass of oxygen produced = mass of mixture – mass of residue

According to the question;

Mass of mixture = 0.950g

Mass of residue = 0.700g

Now, Mass of oxygen produced

= (0.950 - 0.700)(g)

= 0.250(g)

Mass of O₂ = 0.250g

Molecular mass of O₂ = 32g/mol

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of O₂

$$\begin{gathered} =\frac{0.250}{32}\left(mol\right) \\ =0.0078\left(mol\right) \end{gathered}$$

Hence, moles of O₂ produced are 0.0078mol.

According to the balanced equation

$2KCI{O}_3\left(s\right)\stackrel{∆}{→}2KCI\left(s\right)+3O_2\left(g\right)$

2 moles of KClO₃ produces 3 moles O₂

Now, moles of KClO₃

_{localid="1661948307473" $$\begin{gathered} =0.0078×\frac{2}{3}\left(mol\right) \\ =0.0052\left(mol\right) \end{gathered}$$}

Hence, moles of KClO₃ are 0.0052mol.

Moles of KClO₃ = 0.0052mol

Molecular mass of KClO₃ = 122.55g/mol

Again you know,

mass = moles x mass(molar)

Mass of KClO₃

= 0.0052x122.55(g)

= 0.63726(g)

Hence, mass of KClO₃ is 0.63726g.

Again you know,

Mass percent of KClO₃

$$\begin{gathered} =\frac{mass\left(KCI{O}_3\right)}{mass\left(sample\right)}×100\% \\ =\frac{0.63726}{0.950}×100\% \\ =67.08\% \end{gathered}$$

Hence, mass percent of KClO₃ is 67.08%.

Hence, mass percent of KClO₃ in the original mixture is 67.08%.