When heated, a $3.40\ \mathrm g$ sample mixture of $\ce{KClO3}$ decomposes and produces $685\ \mathrm{cm^3}$ $\ce{O2}$ gas at $30\ \mathrm{^\circ C}$ and $101\ \mathrm{kPa}$. What is the mass percentage of $\ce{KClO3}$ in the mixture?
What I've done so far is write out a balanced equation which is:
$$\ce{2KClO3 -> 3O2 + 2KCl}$$
Where do I go from here?
Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematical operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. In the following, we use this method to calculate the amount of oxygen $n_{\ce{O2}}$, the mass of potassium chlorate $m_{\ce{KClO3}}$, and the mass fraction of potassium chlorate $w_{\ce{KClO3}}$ directly from the parameter values given in the question (although only the last calculation is actually required to answer the question).
For simplicity’s sake, you may use the ideal gas law to estimate the amount of oxygen $n_{\ce{O2}}$ as follows. The volume $V=685\ \mathrm{cm^3}=0.000\,685\ \mathrm{m^3}$, the temperature $T=30\ \mathrm{^\circ C}=303.15\ \mathrm K$, and the pressure $p=101\ \mathrm{kPa}=101\,000\ \mathrm{Pa}$ are given in the question. The molar gas constant is $R=8.314\,459\,8(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$.
$$\begin{align} p\cdot V&=n_{\ce{O2}}\cdot R\cdot T\tag1\\[6pt] n_{\ce{O2}}&=\frac{p\cdot V}{R\cdot T}\tag2\\[6pt] &=\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt] &=0.027\,448\,609\ \mathrm{mol}\\[6pt] &\approx0.027\,4\ \mathrm{mol} \end{align}$$
By way of comparison, the amount of oxygen calculated using standard reference data for oxygen instead of using the ideal gas law yields $n_{\ce{O2}}=0.027\,47\ \mathrm{mol}$. Considering the estimated uncertainty of the result obtained by using the ideal gas law, the difference is not significant. Hence, the approximation using the ideal gas law is acceptable in this case.
The chemical equation $$\ce{2KClO3 -> 3O2 + 2KCl}\tag7$$ indicates that $2\ \mathrm{mol}$ of potassium chlorate yield $3\ \mathrm{mol}$ of oxygen. Therefore, $$n_{\ce{KClO3}}=\frac23\cdot n_{\ce{O2}}\tag8$$
The molar mass of potassium chlorate is $M_{\ce{KClO3}}=122.55\ \mathrm{g\ mol^{-1}}$. Since the molar mass $M_{\ce{KClO3}}$ is defined as $$M_{\ce{KClO3}}=\frac {m_{\ce{KClO3}}}{n_{\ce{KClO3}}}\tag9$$ the mass of potassium chlorate $m_{\ce{KClO3}}$ is $$\begin{align} m_{\ce{KClO3}}&=M_{\ce{KClO3}}\cdot n_{\ce{KClO3}}\tag{10}\\[6pt] &=M_{\ce{KClO3}}\cdot \frac23\cdot n_{\ce{O2}}\tag{11}\\[6pt] &=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\tag{12}\\[6pt] &=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt] &=2.242\,551\,324\ \mathrm g\\[6pt] &\approx2.24\ \mathrm g \end{align}$$
The mass fraction of potassium chlorate $w_{\ce{KClO3}}$ is defined as $$w_{\ce{KClO3}}=\frac{m_{\ce{KClO3}}}{m}\tag{13a}$$ where $m=3.40\ \mathrm g$ is the total mass of the sample. Therefore, $$\begin{align} w_{\ce{KClO3}}&=m_{\ce{KClO3}}\cdot\frac1m\tag{13b}\\[6pt] &=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\cdot\frac1m\tag{14}\\[6pt] &=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\times\frac1{3.40\ \mathrm g}\\[6pt] &=0.659\,573\,919\\[6pt] &\approx0.660=66.0\ \% \end{align}$$
