Question

What mass (in g) of oxygen must be added to 9.73 g of neon at 21.9^oC, and in a 82 L contain for a final pressure of 897 mmHg?

Answer 1

step 1: find pressure of Ne

Molar mass of Ne = 20.18 g/mol

mass(Ne)= 9.73 g

use:

number of mol of Ne,

n = mass of Ne/molar mass of Ne

=(9.73 g)/(20.18 g/mol)

= 0.4822 mol

Given:

V = 82.0 L

n = 0.4822 mol

T = 21.9 oC

= (21.9+273) K

= 294.9 K

use:

P * V = n*R*T

P * 82 L = 0.4822 mol* 0.08206 atm.L/mol.K * 294.9 K

P = 0.1423 atm

= 0.1423*760 mmHg

= 108 mmHg

step 2:

find pressure of oxygen

P(O2) = P(total) - P(Ne)

= 897 mmHg - 108 mmHg

= 789 mmHg

step 3:

fins mass of O2

Given:

P = 789.0 mm Hg

= (789.0/760) atm

= 1.0382 atm

V = 82.0 L

T = 21.9 oC

= (21.9+273) K

= 294.9 K

find number of moles using:

P * V = n*R*T

1.0382 atm * 82 L = n * 0.08206 atm.L/mol.K * 294.9 K

n = 3.518 mol

Molar mass of O2 = 32 g/mol

use:

mass of O2,

m = number of mol * molar mass

= 3.518 mol * 32 g/mol

= 113 g

Answer: 113 g