What mass of Oxygen was collected
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What mass of Oxygen was collected

[From: ] [author: ] [Date: 12-07-09] [Hit: ]
How do you get that?n=[(742-24)/760](.0753L)/[(.0821 atm*L/mol*K)(298K)]=.multiply that by atomic mass of O2, which is 32 and you have .......
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?

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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

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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)
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