Question

Small quantities of h2 gas can be collected by adding hcl to zn. a sample of 195 ml of h2 gas was collected over water at 25 c and 753 mm of hg. what mass of h2 was collected? (vp of water at 25 c is 24 mm of hg)

Answer 1

15.4 milligrams The ideal gas law is PV = nRT where P = pressure of the gas V = volume of the gas n = number of moles of gas R = Ideal gas constant (8.3144598 L*kPa/(K*mol) ) T = absolute temperature. So let's determine how many moles of gas has been collected. Converting temperature from C to K 273.15 + 25 = 298.15 K Converting pressure from mmHg to kPa 753 mmHg * 0.133322387415 kPa/mmHg = 100.3917577 kPa Taking idea gas equation and solving for n PV = nRT PV/RT = n n = PV/RT Substituting known values n = PV/RT n = (100.3917577 kPa 0.195 L) / (8.3144598 L*kPa/(K*mol) 298.15 K) n = (19.57639275 L*kPa) / (2478.956189 L*kPa/(mol) ) n = 0.007897031 mol So we have a total of 0.007897031 moles of gas particles. Now let's get rid of that percentage that's water vapor. The percentage of water vapor is the vapor pressure of water divided by the total pressure. So 24/753 = 0.03187251 The portion of hydrogen is 1 minus the portion of water vapor. So 1 - 0.03187251 = 0.96812749 So the number of moles of hydrogen is 0.96812749 * 0.007897031 mol = 0.007645332 mol Now just multiple the number of moles by the molar mass of hydrogen gas. Start with the atomic weight. Atomic weight hydrogen = 1.00794 Molar mass H2 = 1.00794 * 2 = 2.01588 g/mol Mass H2 = 2.01588 g/mol * 0.007645332 mol = 0.015412073 g Rounding to 3 significant figures gives 0.0154 g = 15.4 mg

Answer 2

Answer:- 0.0153 grams

Explanation:- According to Dalton's law, the total pressure is the sum of individual pressures.

`p_(total)=p_(H_2O)+p{H_2}`

`p753mmHg=24mmHg+p{H_2}`

`p_(H_2)=729mmHg`

According to the ideal gas equation:'

`PV=nRT`

Pressure of the gas = 729 mmHg = 0.96 atm (1mmHg=0.0013atm)

Volume of the gas = 195 mL = 0.195 L

Temperature of the gas = 25°C =(25+273)= 295K (0°C = 273 K)

R= gas constant = 0.0821Latm\Kmol

`n=(PV)/(RT)=(0.96* 0.195 )/(0.0821 * 298 K)=7.65* 10^(-3)mol`

Mass of
`H_2=moles* {\text {Molar mass}}=7.65* 10^(-3)* 2=0.0153g`