Question

Question

Answer 1

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

Answer 2

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