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Answer :
To produce 38.2 L of oxygen gas at a pressure of 2.00 atm and a temperature of 310 K, approximately 96.27 grams of potassium chlorate (KClO₃) are needed.
To determine the mass of potassium chlorate needed, we can use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
First, we need to calculate the number of moles of oxygen gas using the ideal gas law equation. Rearranging the equation to solve for n, we have n = PV / RT. Plugging in the values, n = (2.00 atm) * (38.2 L) / (0.0821 L·atm/(mol·K) * 310 K), we find n ≈ 2.374 moles.
The balanced chemical equation for the decomposition of potassium chlorate into oxygen gas is 2KClO₃ → 2KCl + 3Oâ‚‚. According to the equation, 2 moles of potassium chlorate produce 3 moles of oxygen gas. Therefore, the number of moles of potassium chlorate required is (2.374 moles Oâ‚‚) / (3 moles KClO₃) ≈ 0.791 moles.
To calculate the mass of potassium chlorate, we can use the molar mass of KClO₃, which is approximately 122.55 g/mol. The mass of potassium chlorate needed is (0.791 moles) * (122.55 g/mol) ≈ 96.27 grams. Therefore, approximately 96.27 grams of potassium chlorate are needed to produce 38.2 L of oxygen gas at a pressure of 2.00 atm and a temperature of 310 K.
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