Question

What volume of oxygen gas in L at 426.15 K and 0.820 atm can be produced by the decomposition of 22.4 g of solid KClO3 according to the balanced equation below?

2KClO3 (s) → KCl (s) + 3O2 (g)

Answer 1

To find the volume of oxygen gas produced by the decomposition of 22.4 g of KClO3, calculate the moles of KClO3, determine the moles of O2 produced from the balanced chemical equation, and use the ideal gas law to calculate the volume of O2 gas.

Step-by-step explanation:

The question involves calculating the volume of oxygen gas produced by the decomposition of potassium chlorate (KClO3) using the given balanced chemical equation and the conditions of temperature (426.15 K) and pressure (0.820 atm). To solve this, we first need to find the number of moles of KClO3 that corresponds to 22.4 g using the molar mass of KClO3 (122.55 g/mol). Once we have the moles of KClO3, we can then determine the moles of oxygen gas produced using the stoichiometry of the balanced equation. Finally, we apply the ideal gas law to calculate the volume of oxygen gas.

Steps to Calculate Volume of Oxygen Gas

1. Convert the mass of KClO3 to moles: moles KClO3 = mass KClO3 / molar mass KClO3.
2. Use the stoichiometry of the reaction to find the moles of O2 produced: moles O2 = (3/2) × moles KClO3.
3. Apply the ideal gas law: V = (nRT) / P, where V is the volume of O2, n is the moles of O2, R is the ideal gas constant (0.08206 L·atm/K·mol), T is the temperature in Kelvin, and P is the pressure in atm.