The reaction of 0.194 g of Zn with HCl_(aq) gave 80.3 mL (at 23.4 degree C) of h
ID: 515841 • Letter: T
Question
The reaction of 0.194 g of Zn with HCl_(aq) gave 80.3 mL (at 23.4 degree C) of hydrogen gas collected over water in a gas tube. The water was 18.5 mm higher in the gas tube than it was in the beaker (step C). The barometer reading is 698. mm Hg. Following the steps on pages 1 to 3, complete this table of data. A. Mass of Zn B. Volume of H_2 Gas Generated C. Height of water Column inside tube (mm H_2 O) D. Temperature of H_2 Gas E. Vapor Pressure of Water F. Atmospheric Pressure (mm Hg) G. Adjustment for water in tube: "C." times 1 mm Hg/13.5 mm H_2 O H. Total Pressure inside gas tube (mm Hg) (F-G) I. Pressure of hydrogen inside gas tub (mm Hg) (H - E) J. Pressure of hydrogen in atm K. Moles of Hydrogen from PV = nRT L. Balanced Chemical Equation M. Calculated Molar Mass of ZincExplanation / Answer
A. Mass of Zn: 65.37 g/mol
B. Volume of H2 gas generated 80.3 ml
C. Height of water column inside tube 18.5 mm
D. Temerature of H2 gas 23.4 oC
273.15+23.4= 296.55 K
E. Vapor pressure of water 716.5 mm?
F. Atmospheric pressure (mm Hg) 690 mm
G. Adjustment for water in tube 'C' 18.5 mm x 1mm/ 13.5 mm H2O 1.39 mm?
H. Total pressure inside gas tube 690 mm + 1.39 mm 691.39 mm?
I. Pressure of hydrogen inside gas tube
J. Pressure of hydrogen in atm
K. Moles of hydrogen from PV =nRT
Or it should be equal to no. of moles of Zinc used in the reaction according to balanced equation = 0.00296 mol
L. Balanced chemical equation:
Zn (s) + 2 HCl (aq) ------------------------> H2 (g) + ZnCl2 (aq)
M. Calculated molar mass of Zinc 0.194 g / 65.37 g = 0.00296 moles or 2.96 mmoles
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.