how to calculate the pressure of H2 and moles of H2 Experiment 10 The Ideal Gas
ID: 561058 • Letter: H
Question
how to calculate the pressure of H2 and moles of H2
Experiment 10 The Ideal Gas Law Constant (R) Name Data Table Usc the following table to record your data for cach trial Trial 1 Trial 2 Trial 3 Atmospheric pressure Mass of magnesium metal Volumc of 6 M hydrochloric acid Volume of molecular hydrogen gas Water temperature Partial pressure of water at above gm 2 21.3 2450 2 temperature) Calculations Do the following calculations for each trial Pressure of Hgas: Assume that water and molecular hydrogen are the only gases inside the tube. Using the measured atmospheric pressure and the partial pressure of water vapor (determined fronm the temperature see Appendix 3), calculate the pressure of molecular hydrogen gas for each trial 4 ,239k Moles of H, gas: Using the molar mass of magncsium and the balanced equation for the reaction you have just observed, calculate the moles of molecular hydrogen that should have been formed This calculation assumes that the magnesium was pure and was completely consumed by the hydrochloric acid, and that no H, was lost in the process (i.e., it assumes a 100% yield of hydrogen gas).Explanation / Answer
Pressure of H2 gas:
Trial 1
Trial 2
Atmospheric pressure (mmHg)
52
53
Partial pressure of water (mmHg)
18.650
18.996
Partial pressure of H2 gas = (atmospheric pressure) – (partial pressure of water) (mmHg)
52 – 18.650 = 31.350
53 – 18.996 = 34.004
Partial pressure of H2 gas in atmospheres = (partial pressure of H2 gas in mmHg)*(1 atm/760 mmHg) (atm)
(31.350)*(1/760) = 0.04125
(34.004)*(1/760) = 0.04474
Moles of H2 gas:
The balanced chemical equation for the reaction between Mg and HCl is
Mg (s) + 2 HCl (aq) --------> MgCl2 (aq) + H2 (g)
As per the stoichiometric equation,
1 mole Mg = 1 mole H2
Therefore, mole(s) of Mg used in the reaction = mole(s) of H2 produced.
We will require the atomic mass of Mg which is 24.305 g/mol.
Trial 1
Trail 2
Mass of Mg used (g)
0.082
0.080
Moles of Mg used = (mass of Mg used)/(atomic mass of Mg)
(0.082)/(24.305) = 0.003374
(0.080)/(24.305) = 0.003291
Moles of H2 produced = moles of Mg used
0.003374
0.003291
Trial 1
Trial 2
Atmospheric pressure (mmHg)
52
53
Partial pressure of water (mmHg)
18.650
18.996
Partial pressure of H2 gas = (atmospheric pressure) – (partial pressure of water) (mmHg)
52 – 18.650 = 31.350
53 – 18.996 = 34.004
Partial pressure of H2 gas in atmospheres = (partial pressure of H2 gas in mmHg)*(1 atm/760 mmHg) (atm)
(31.350)*(1/760) = 0.04125
(34.004)*(1/760) = 0.04474
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