Question

small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, kclo₃(s). the equation for the reaction is 2kclo₃ → 2kcl + 3o₂. calculate how many grams of o₂(g) can be produced from heating 28.2 g kclo₃(s).

Explanation:

Step1: Calculate molar masses

The molar mass of $KClO_3$: $M_{KClO_3}=39.1 + 35.5+3\times16=122.6\ g/mol$. The molar mass of $O_2$ is $M_{O_2}=2\times16 = 32\ g/mol$.

Step2: Determine mole - ratio

From the balanced chemical equation $2KClO_3
ightarrow2KCl + 3O_2$, the mole - ratio of $KClO_3$ to $O_2$ is $2:3$.

Step3: Calculate moles of $KClO_3$

The number of moles of $KClO_3$, $n_{KClO_3}=\frac{m_{KClO_3}}{M_{KClO_3}}=\frac{28.2\ g}{122.6\ g/mol}\approx0.23\ mol$.

Step4: Calculate moles of $O_2$

Using the mole - ratio, if $n_{KClO_3} = 0.23\ mol$, then $n_{O_2}=\frac{3}{2}n_{KClO_3}=\frac{3}{2}\times0.23\ mol = 0.345\ mol$.

Step5: Calculate mass of $O_2$

The mass of $O_2$, $m_{O_2}=n_{O_2}\times M_{O_2}=0.345\ mol\times32\ g/mol = 11.04\ g$.

Answer:

$11.04\ g$