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tony3320 tony3320
wrote...
11 years ago
Hydrogen gas is produced by the reaction of hydrochloric acid on zinc metal.  The gas is collected over water.  If 156 .0 ml of gas is collected at 19.0 degrees Celsius and 760.0 mm Hg, what is the mass of hydrogen collected?
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SMOKEY2112SMOKEY2112
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11 years ago
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michelecul Author
wrote...
11 years ago
Calculate volume at STP = 0°C or 273K , and 1 atm pressure. = 760mmHg.
V1/T1 = V2/T2
V1/273 = 156/292
V1 = 156*273/292
V1 = 145.85ml

At STP, 1mol has volume = 22.40 litres
 22.4 litres has mass 2.016g
0.14585 litres has mass =  0.14585/22.4*2.016 = 1.31*10^-2g H2 gas produced.
wrote...
11 years ago
when you collect a gas over water the gas mixture will contain water vapour. You need to determine the actual partial pressure of the collected gas in the mixture of gas and water vapour..

Partial pressure H2 = total pressure - vapour pressure H2 at 19 deg C
= 760.0 mmHg - 16.5 mmHg
= 743.5 mmHg
http://intro.chem.okstate.edu/1515SP01/Database/VPWater.html

Then use ideal gas equaion to determine moles of H2

PV = nRT

P = pressure H2 = 743.5 mmHg
V = volume = 156.0 ml = 0.1560 L
n = moles = ?
R = gas constant = 62.36367 mmHg L mol^-1 K^-1
http://en.wikipedia.org/wiki/Gas_constant
T = temp in Kelvin = 292 K

n = PV / RT
= 743.5 mmHg x 0.1560 L / 62.36367 x 292 K
= 0.0063693 moles H2

mass = molar mass x moles
= 2.016 g/mol x 0.0063693 mol
= 0.01284 g
= 1.28 x 10^-2 g
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