Question

In the reaction: \[ 2 \text{KClO}_3 (s) \rightarrow 2 \text{KCl} (s) + 3 \text{O}_2 (g) \quad \Delta H = -78 \, \text{kJ} \] If 33.6 L of oxygen gas is liberated at S.T.P., what is the mass of KClO\(_3\) produced?

• A. 48.0 g
• B. 7.45 g
• C. 24.0 g
• D. 74.5 g

Hint:

When dealing with reactions involving gases, use stoichiometry and the ideal gas law to convert volumes of gases to moles and then to the corresponding mass of a solid.

Answer 1

The Correct Option is D

Step 1: Use the Ideal Gas Law and Stoichiometry.
From the reaction, 2 moles of KClO\(_3\) produce 3 moles of O\(_2\). At S.T.P., 1 mole of gas occupies 22.4 L. So, the moles of O\(_2\) liberated are: \[ \frac{33.6 \, \text{L}}{22.4 \, \text{L/mol}} = 1.5 \, \text{mol O}_2 \]
Step 2: Relate moles of O\(_2\) to moles of KClO\(_3\).
From the stoichiometry of the reaction: \[ \frac{2 \, \text{mol KClO}_3}{3 \, \text{mol O}_2} \times 1.5 \, \text{mol O}_2 = 1.0 \, \text{mol KClO}_3 \]
Step 3: Calculate the mass of KClO\(_3\).
The molar mass of KClO\(_3\) is: \[ \text{K} = 39, \, \text{Cl} = 35.5, \, \text{O} = 16 \quad \Rightarrow \, \text{Molar mass of KClO}_3 = 122.5 \, \text{g/mol} \] Thus, the mass of KClO\(_3\) is: \[ 1.0 \, \text{mol} \times 122.5 \, \text{g/mol} = 74.5 \, \text{g} \]
Step 4: Conclusion.
The mass of KClO\(_3\) produced is 74.5 g, which corresponds to option (D).