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 95.9 grams of KClO3(s).

Answer 1

Answer on Question #40476 - Chemistry - Other

Question

Small quantities of oxygen can be prepared in the laboratory by heating potassium chlorate, $\mathrm{KClO}_3(\mathrm{s})$. The equation for the reaction is

$$2 \mathrm{KClO}_3 \rightarrow 2 \mathrm{KCl} + 3 \mathrm{O}_2$$

Calculate how many grams of $\mathrm{O}_2(\mathrm{g})$ can be produced from heating 95.9 grams of $\mathrm{KClO}_3(\mathrm{s})$.

Answer:

Molar mass of $\mathrm{KClO}_3$ equals:

$$M(\mathrm{KClO}_3) = M(K) + M(\mathrm{Cl}) + 3M(\mathrm{O}) = 39.1 + 35.5 + 3 \cdot 16.0 = 122.6 \frac{\mathrm{g}}{\mathrm{mole}}$$

Mass of 2 moles of potassium chlorate equals:

$$2 \cdot 122.6 = 245.2 \mathrm{g}$$

Molar mass of $\mathrm{O}_2$ equals:

$$M(\mathrm{O}_2) = 2M(\mathrm{O}) = 2 \cdot 16.0 = 32.0 \frac{\mathrm{g}}{\mathrm{mole}}$$

Mass of 3 moles of $\mathrm{O}_2$ equals:

$$3 \cdot 32.0 = 96 \mathrm{g}$$

Then we make a proportion:

$$\begin{array}{l} 245.2 \mathrm{g} \text{ of } \mathrm{KClO}_3 \text{ produce } 96.0 \mathrm{g} \text{ of } \mathrm{O}_2 \\ 95.9 \mathrm{g} \text{ of } \mathrm{KClO}_3 - x \mathrm{g} \text{ of } \mathrm{O}_2 \\ x = \frac{95.9 \cdot 96.0}{245.2} = 37.5 \mathrm{g} \\ \end{array}$$

Answer: $m(\mathrm{O}_2) = 37.5 \mathrm{g}$.