Question

Question

14

of 18

7 points

How many liters are occupied by 18.0g of oxygen gas (O2)at STP (standard air

temperature and pressure)? ONLY ENTER THE NUMBER OF YOUR

ANSWER. ROUND TO THE HUNDREDTHS PLACE IN YOUR ANSWER. DO

NOT LABEL

Answer:

Submit

Answer 1

The volume of oxygen gas at STP is 12.60 L

Step-by-step 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

12.62 L

Step-by-step 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