Volume pressure= .785 liters Unknown 1 = 1884 torr Unknown 2 = 847 torr Now that

Question

Volume pressure= .785 liters

Unknown 1 = 1884 torr

Unknown 2 = 847 torr

Now that the volume of the pressure vessel is known, the number of moles H2 gas formed can be determined for the unknown samples. Enter the corrected gas pressure after subtracting the water vapor pressure for unknowns 1 and 2

Unknown 1 corrected pressure?

Unknown 2 corrected pressure?

Use the ideal gas equation and the volume obtained above to calculate the number of moles H2 gas produced during each unknown run. Assume again that the gas temperature at the end of each run is 22 degrees C

Unknown 1?

Unknown 2?

Divide the sample mass by the number of moles of H2, calculated above to determine the ratio of sample mass to number of moles of H2, produced.

Explanation / Answer

vapor pressure of water at 22 deg.c= 19.8 mm Hg

volume unknonw= 0.785 Liters

Unkown 1 coreected pressure = total pressure -19.8 =1884 mm-19,8mm ( since 1 Torr = 1mm) =1864.2 mmHg =1864.2/760 =2.452 atm

P= 2.452 atm T= 22 deg.c= 22+273.15= 295.15K V= 0.785 L, R =0.08206L.atm/mole.K

number of moles, n =PV/RT = 2.452*0.785/(0.08206*295.15)=0.079 moles of H2

b) Unknown 2 corrected pressure, P=847- 19.8 =827.2 mm Hg= 827.2/760=1.088 atm

n= 1.088*0.785/ (0.08206*295.15)=0.035263 moles

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