Hydrogen peroxide was catalytically decomposed and 75.3 mL of oxygen gas was collected over water at 25°C and 742 torr. What mass (not moles) of oxygen was collected? (Pwater = 24 torr at 25°C)
The answer is .0931 grams. How do you get that?
The answer is .0931 grams. How do you get that?
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Use PV=nRT to find moles so
n=[(742-24)/760](.0753L)/[(.0821 atm*L/mol*K)(298K)]=.002908 moles
multiply that by atomic mass of O2, which is 32 and you have .0931 g
n=[(742-24)/760](.0753L)/[(.0821 atm*L/mol*K)(298K)]=.002908 moles
multiply that by atomic mass of O2, which is 32 and you have .0931 g
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Use PV=nRT
where
P= 742torr = 98925Pa
V=75.3mL = 0.0000753m3
R= gas constant = 8.314 J/K mol
T = 25C = 298K
n = PV/RT = (98925x0.000075)/(8.314x298) = 0.003 mols
mass = n x RMM = 0.003 x 32 = 0.096g (close enough)
where
P= 742torr = 98925Pa
V=75.3mL = 0.0000753m3
R= gas constant = 8.314 J/K mol
T = 25C = 298K
n = PV/RT = (98925x0.000075)/(8.314x298) = 0.003 mols
mass = n x RMM = 0.003 x 32 = 0.096g (close enough)