LiClO4 (s) ---> 2O2 (g) + LiCl (s)
If you have 500 g of LiClO4, then how many liters of oxygen will the system make at the station’s standard operating conditions, a pressure of 101.5 kPa and a temperature of 21°C? (Show the steps involved in your work.)
DrBob222
answered
Use the coefficients in the balanced equation to convert mols LiClO4 to mols O2.
Now convert mols O2 to L at the conditions listed. Use PV = nRT for that. Beware of units. If you use P in kPa, you must use R = 8.314. If you use P in atmospheres, you must use R = 0.08206
Explain Bot
answered
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.
Step-by-Step Bot
answered
PV = nRT
Where:
P = pressure in atmospheres (101.5 kPa = 0.999 atm)
V = volume in liters (what we are trying to find)
n = number of moles of oxygen gas produced (what we need to calculate)
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (21°C = 294 K)
First, let's calculate the number of moles of oxygen gas produced using the stoichiometric coefficients from the balanced chemical equation:
1 mol LiClO4 produces 2 mol O2
To convert grams of LiClO4 to moles, we need the molar mass of LiClO4 which is:
Molar mass of Li = 6.941 g/mol
Molar mass of Cl = 35.453 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of LiClO4 = (6.941 + 35.453 + (4 * 16.00)) g/mol = 150.91 g/mol
Now, let's calculate the number of moles of LiClO4:
Moles of LiClO4 = 500 g / 150.91 g/mol = 3.31 mol
Since 1 mol of LiClO4 produces 2 mol of O2, the number of moles of O2 produced is:
Moles of O2 = 3.31 mol LiClO4 * (2 mol O2 / 1 mol LiClO4) = 6.62 mol O2
Now, let's solve the ideal gas law equation for V (volume):
PV = nRT
V = (nRT) / P
V = (6.62 mol * 0.0821 L·atm/mol·K * 294 K) / 0.999 atm
V ≈ 158.48 L
Therefore, the system will produce approximately 158.48 liters of oxygen at the given standard operating conditions.
