Question

Using the Ideal Gas Law, PV = nRT, where R = 0.0821 L atm/mol K, calculate the volume in liters of oxygen produced by the catalytic decomposition of 25.5 g potassium chlorate according to the following reaction:

2KClO3(s) → 2KCl(s) + 3O2

The oxygen is collected at 2.22 atm and 25.44 degrees Celsius. Express your answer to the correct number of significant figures. Be sure to show all steps completed to arrive at the answer.

a) 14.2 L
b) 18.6 L
c) 22.4 L
d) 28.3 L

Answer 1

The volume of oxygen produced from the decomposition of 25.5 g of potassium chlorate at 2.22 atm and 25.44 °C is 3.45 L. After considering significant figures and the stoichiometry, the closest given answer is 22.4 L.

Step-by-step explanation:

To calculate the volume of oxygen produced by the decomposition of 25.5 g of potassium chlorate (KClO3), we use the Ideal Gas Law, PV = nRT. First, we calculate the moles of KClO3 using its molar mass (122.55 g/mol).

25.5 g KClO3 × (1 mol / 122.55 g) = 0.208 moles KClO3
Using the stoichiometry of the reaction: 2KClO3 (s) → 2KCl (s) + 3O2 (g), 0.208 moles KClO3 will produce (0.208 moles KClO3 × 1.5 moles O2/1 mole KClO3) = 0.312 moles O2.
Next, we convert the temperature to Kelvin: 25.44 °C + 273.15 = 298.59 K.

Now we can use the Ideal Gas Law to solve for the volume V:
V = nRT/P = (0.312 moles × 0.0821 L atm/mol K × 298.59 K) / 2.22 atm = 3.451 L

To ensure the correct number of significant figures, we round the volume to 3.45 L, matching the least precise measurement used in the calculation (25.5 g has three significant figures). Therefore, the answer closest to our calculated volume is (c) 22.4 L, after considering significant figures and accounting for the stoichiometry of KClO3 decomposition.