The molar volume of H2 gas at STP is 22.433 L/mol. We want to verify this number

Question

The molar volume of H2 gas at STP is 22.433 L/mol. We want to verify this number in this experiment by reacting a strong acid with a metal to produce hydrogen gas. The following reaction takes place in a eudiometer tube: 2H+(aq) + Mg(s) Mg2+(aq) + H2 (g) A0.0302 g sample of elemental magnesium is reacted with an abundant amount of hydrochloric acid> The volume of hydrogen gas collected over water is 33.44 mL temperature of the water is 23.2 °C. (Refer to Table 9.1 for water vapor pressure values.) The atmospheric pressure recorded in lab is 748.2 torr. The pressure difference between the two menisci is 460.0 mm H2O, Calculate the following values in the unit required by the relevant formulas and show all your work Calculation Molar mass of Mg = (1 point) Atmospheric pressure in hspPressn or 7 mm Hg = (1 point) 74f.eturn.I7wanm H Su «V:~2ony,,. Water vapor pressure in mm Hg = (1 point) … 3.3 z/,omt 3 Pressure difference in mm Hg = (1 point)

Explanation / Answer

Molar mass of magnesium =24 g/mol

Atmospheric pressure = 748.2 torr= 748.2 mm Hg ( 1 Torr =1mm Hg)

Vapor pressure of water at 23.2 deg.c= 21.329 mm Hg

Pressure difference= 460mm H2O

10336 mm water =760 mm Hg

460mm water= 460*760/10336=33.82 mm Hg

Partial pressure of H2 Gas= total pressure- saturation vapor pressure of water- Pressure difference= 748.2-21.329-33.82=693.051 mm Hg

Water bath temperature = 23.2 deg.c= 23.2+273.15=296.35 K

Volume of hydrogen collected= 33.44 ml=0.03344 L

V1= 0.03344 L, P= 693.051mm T= 296.35 K

V2=? P2= 760mm Hg T2= 273.15 ( STP conditions)

From P1V1/T1= P2V2/T2   and V2= P1V1*T2/(T1*P2)= 693.05*0.03344*273.15/(296.35*760)=0.028107 L

Moles of Mg= Mass/ atomic weight = 0.0302/24=0.00126 moles

Moles of hydrogen =0.00126 ( 1 mole of Mg gives 1 mole of H2)

Molar volume of hydrogen = 0.028107/0.00126=22.31 L

% error =100* (22.4-22.31)/22.4 =0.4%

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