To calculate the volume of oxygen gas produced, 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 ideal gas constant, and T is the temperature.
Given:
Pressure, P = 101.5 kPa
Temperature, T = 21°C = 294 K
First, let's calculate the number of moles of oxygen produced by reacting 500 g of LiClO4 using stoichiometry.
1. Calculate the molar mass of LiClO4:
Molar mass of Li = 6.94 g/mol
Molar mass of Cl = 35.45 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of LiClO4 = (6.94 + 35.45 + (16.00 * 4)) g/mol = 117.37 g/mol
2. Calculate the number of moles of LiClO4:
Moles of LiClO4 = Mass of LiClO4 / Molar mass of LiClO4
= 500 g / 117.37 g/mol
= 4.26 mol
3. According to the balanced chemical equation, 1 mole of LiClO4 yields 2 moles of O2. Therefore, the number of moles of O2 produced is twice the moles of LiClO4 used.
Moles of O2 = 2 * Moles of LiClO4
= 2 * 4.26 mol
= 8.52 mol
Now, let's calculate the volume of oxygen gas produced using the ideal gas law equation:
4. Convert the pressure from kPa to atm:
Pressure = 101.3 kPa * 1 atm / 101.3 kPa
= 1 atm
5. Calculate the volume of oxygen gas:
V = nRT / P
= (8.52 mol) * (0.0821 L∙atm/mol∙K) * (294 K) / (1 atm)
≈ 218 L
Therefore, approximately 218 liters of oxygen gas will be produced at the station's standard operating conditions.