Question

How many pounds of KClO₃ must used to obtain 245 lb of oxygen?

Answer 1

To determine how many pounds of potassium chlorate (KClO₃) are needed to obtain 245 pounds of oxygen (O₂), we first need to look at the chemical reaction involved in the decomposition of potassium chlorate:

\[
2 \, \text{KClO}_3(s) \rightarrow 2 \, \text{KCl}(s) + 3 \, \text{O}_2(g)
\]

From the reaction, we can see that:

- 2 moles of KClO₃ produce 3 moles of O₂.

Next, we need to calculate the molar masses to relate the amount of KClO₃ to the amount of oxygen produced:

- The molar mass of KClO₃ is:
- K: 39.10 g/mol
- Cl: 35.45 g/mol
- O₃: 3 × 16.00 g/mol = 48.00 g/mol
- Total = 39.10 + 35.45 + 48.00 = 122.55 g/mol

- The molar mass of O₂ is:
- O₂: 2 × 16.00 g/mol = 32.00 g/mol

Now, from the stoichiometry of the reaction, we can set up the proportion:

\[
\frac{2 \, \text{moles KClO}_3}{3 \, \text{moles O}_2}
\]

We want to find how many moles of KClO₃ are needed to produce 245 pounds of O₂.

1. Convert pounds of O₂ to grams:
\[
245 \, \text{lb} \times 453.592 \, \text{g/lb} = 111,130.84 \, \text{g}
\]

2. Calculate the number of moles of O₂ produced:
\[
\text{Moles of O}_2 = \frac{111,130.84 \, \text{g}}{32.00 \, \text{g/mol}} \approx 3,470.35 \, \text{mol}
\]

3. Using the stoichiometric ratio, calculate the moles of KClO₃ required:
\[
\text{Moles of KClO}_3 = \left( \frac{2 \, \text{moles KClO}_3}{3 \, \text{moles O}_2} \right) \times 3,470.35 \, \text{mol} \approx 2,313.57 \, \text{mol}
\]

4. Now, convert moles of KClO₃ to grams:
\[
\text{Mass of KClO}_3 = 2,313.57 \, \text{mol} \times 122.55 \, \text{g/mol} \approx 283,400.8 \, \text{g}
\]

5. Convert grams of KClO₃ to pounds:
\[
\text{Mass of KClO}_3 \, (\text{lb}) = \frac{283,400.8 \, \text{g}}{453.592 \, \text{g/lb}} \approx 624.0 \, \text{lb}
\]

Therefore, to obtain 245 pounds of oxygen, you would need approximately **624 pounds of KClO₃**.