Answer:
12.62 L
Explanation:
First, we have to calculate the moles corresponding to 18.0 g of oxygen gas (MW 32.0).
18.0 g × (1 mol/32.0 g) = 0.563 mol
Then, we can find the volume occupied by 0.563 moles of oxygen at STP (273,15 K, 1.00 atm) using the ideal gas law.
P × V = n × R × T
V = n × R × T / P
V = 0.563 mol × 0.0821 atm.L/mol.K × 273.15 K / 1.00 atm
V = 12.62 L
Answer:
The volume of oxygen gas at STP is 12.60 L
Explanation:
Step 1: data given
Mass of oxygen = 18.0 grams
Molar mass of oxygen = 32.0 g/mol
STp = 1atm and 273 K
Step 2: Calculate moles of oxygen
Moles O2 = mass O2 / molar mass O2
Moles O2 = 18.0 grams / 32.0 g/mol
Moles O2 = 0.5625 moles
Step 3: Calculate volume of the gas
p*V = n*R*T
⇒with p = the pressure at STP = 1.00 atm
⇒with V = the volume of the oxygen gas = TO BE DETERMINED
⇒with n = the number of moles of oxygen gas = 0.5625 moles
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature at STP = 273 K
V = (n*R*T)/p
V = (0.5625*0.08206 * 273) / 1 atm
V = 12.60 L
The volume of oxygen gas at STP is 12.60 L
