help with chemistry question

Question

A 10.0 L reaction vessel captured oxygen gas, O2, and registered a pressure of 152 kPa at a temperature of 0 °C. Find the mass of the oxygen gas.

Answer 1

Ideal Gas Law:

PV = nRT

 

R is a constant, choose the correct one for the units you're using. Or convert units to match R.

 

V = 10 L

P = 152kPa * 0.00986923 atm/kPa = 1.5 atm

T = 0°C + 273.15 K = 273.15 K

R = 0.08206 L atm mol^¯1 K^¯1

 

(1.5 atm)*(10 L) = n( 0.08206 L atm mol^¯1 K^¯1)*(273.15 K)

 

n = 0.67 mole O₂

 

Molecular weight of O₂ = 16 + 16 = 32 g/mol

 

0.67 mol O₂ * 32 g/mol = 21.42 g O₂

 

Answer 2

From PV = nRT you know

V = 10.0 L

P = 152 kPa

R = gas constant (varies depending on what units you use)

T = 0ºC = 273ºK

n = ?  You solve for n (moles of O₂)

n = PV/RT 

if you use 152 kPa as the pressure, you need to use R = 8.314 L-kPa/mol-K

n = (152)(10)/(8.314)(273)

n = 0.6697 moles O₂

0.6697 moles O₂ x 32 g/mole = 21.4 g O₂