Question

Calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate. 2KClO3 --> 2KCl + 3O2.
Responses

29.5 g
29.5 g

0.922 g
0.922 g

19.67 g
19.67 g

112.5 g

Answer 1

To calculate the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate \((KClO_3)\), we first need to determine the molar mass of potassium chlorate.

1. **Calculate the molar mass of \(KClO_3\)**:
- Potassium (K): 39.10 g/mol
- Chlorine (Cl): 35.45 g/mol
- Oxygen (O): 16.00 g/mol

\[
\text{Molar mass of } KClO_3 = 39.10 + 35.45 + (3 \times 16.00) = 39.10 + 35.45 + 48.00 = 122.55 \text{ g/mol}
\]

2. **Convert grams of \(KClO_3\) to moles**:
\[
\text{Moles of } KClO_3 = \frac{75.0 \text{ g}}{122.55 \text{ g/mol}} \approx 0.612 \text{ mol}
\]

3. **Use the stoichiometry of the reaction**:
The balanced equation is:
\[
2KClO_3 \rightarrow 2KCl + 3O_2
\]

From the balanced equation, 2 moles of \(KClO_3\) produce 3 moles of \(O_2\). Therefore, we can find the moles of \(O_2\) produced from the moles of \(KClO_3\):
\[
\text{Moles of } O_2 = 0.612 \text{ mol } KClO_3 \times \frac{3 \text{ mol } O_2}{2 \text{ mol } KClO_3} = 0.918 \text{ mol } O_2
\]

4. **Convert moles of \(O_2\) to grams**:
The molar mass of \(O_2\) is:
\[
32.00 \text{ g/mol}
\]
\[
\text{Mass of } O_2 = 0.918 \text{ mol} \times 32.00 \text{ g/mol} \approx 29.376 \text{ g}
\]

5. **Final result**:
Rounding this to three significant figures gives approximately:
\[
29.4 \text{ g}
\]
or when rounded appropriately:
\[
\approx 29.5 \text{ g}
\]

Thus, the mass of oxygen produced from the decomposition of 75.0 g of potassium chlorate is **29.5 g**.