The reaction of 54.3 mg of Al with HCl (aq) gave 86.2 mL (at 25.5°C) of hydrogen

Question

The reaction of 54.3 mg of Al with HCl(aq) gave 86.2 mL (at 25.5°C) of hydrogen gas collected over water in a gas tube. The water was 8.5 mm higher in the gas tube than it was in the beaker. The uncorrected barometer reading is 702.3 mm Hg at 23.0°C. Complete this table of data.

A. Mass of Al   B. Volume of H2 Gas Generated   C. Height of Water Column inside tube (mm H2O)   D. Temperature of H2 Gas

E. Vapor Pressure of Water    F. Uncorrected Atmospheric Pressure (mm Hg)   

G. Corrected Atmospheric Pressure (mm Hg)   H. "C." x 1 mm Hg/13.5 mm H2O    

I. Total Pressure inside gas tube (mm Hg) (G-H)

J. Pressure of hydrogen inside gas tube (mm Hg) (I - E) K. Pressure of hydrogen in atm  

L. Moles of Hydrogen from PV = nRT      M. Balanced Chemical Equation         

Explanation / Answer

A: Mass of Al = 54.3 mg

B: Volume of H2 gas generated = 86.2 mL

C): Height of Water Column inside tube (mm H2O) = 8.5 mm

D. Temperature of H2 Gas = 25.5 DegC = 25.5 + 273 = 298.5 K

E. Vapor Pressure of Water = 24.4 mm Hg

F. Uncorrected Atmospheric Pressure (mm Hg) = 702.3 mm Hg

G. Corrected Atmospheric Pressure (mm Hg) = 705.1 mm Hg

H. "C." x 1 mm Hg/13.5 mm H2O = 8.5 mm H2O x ( 1 mm Hg / 13.5 mm H2O) = 0.630 mm Hg

I. Total Pressure inside gas tube (mm Hg) (G-H) = 705.1 mm Hg - 0.630 mm Hg = 704.47 mm Hg

J. Pressure of hydrogen inside gas tube (mm Hg) (I - E) = 704.47 mm Hg - 24.4 mm Hg = 680.07 mm Hg

K. Pressure of hydrogen in atm = 0.894830 atm

L. Moles of Hydrogen from PV = nRT     

=> n = PV / RT = (0.894830 atm x 0.0862 L) / (0.0821 L.atm.mol-1K-1 x 298.5K) = 0.00315 mol

M. Balanced Chemical Equation:

2 Al(s) + 6HCl(aq) ------- > 2 AlCl3 + 3H2(g)

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