1.) A sample of gaseous hydrogen gas was collected by the displacement of water

Question

1.) A sample of gaseous hydrogen gas was collected by the displacement of water at 21.8°C and a barometric pressure of 752 torr. The volume under these conditions was measured to be 35.76 mL. The gas was generated by the reaction of a sample of metal weighing 15.45 grams and containing magnesium metal. The reaction was carried out with with HCI as Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g) a.) How many moles of hydrogen gas were produced by this sample? (See Lab Notebook pages 10-12) b) Calculate the % m/m of magnesium in the sample.

Explanation / Answer

1) sample of metal Mg was reacted with HCl

Mg + 2HCl --> MgCl2 + H2(g)

a) water vapor pressure at 21.8 oC = 19.6 torr

Pressure of H2 gas collected = 752 - 19.6 = 732.4 torr

pressure of H2 gas in atm (P) = 732.4/760 = 0.964 atm

Volume (T) = 35.76 ml/1000 = 0.03576 L

Temperature (T) = 21.8 oC + 273 = 294.8 K

Gas constant (R) = 0.08205 L.atm/K.mol

Using ideal gas equation,

moles (n) of H2 gas = PV/RT

                                 = 0.964 x 0.03576/0.08205 x 294.8

                                 = 0.00143 mole

b) moles of Mg reacted in the sample = 0.00143 mol

mass of Mg actual present in sample = 0.00143 mol x 24.305 g/mol = 0.035 g

% m/m Mg/metal in sample = 0.035 g x 100/15.45 g = 0.23%

c) If water vapor pressure was not taken into consideration,

pressure of H2 gas in atm (P) = 752/760 = 0.99 atm

Volume (T) = 35.76 ml/1000 = 0.03576 L

Temperature (T) = 21.8 oC + 273 = 294.8 K

Gas constant (R) = 0.08205 L.atm/K.mol

Using ideal gas equation,

moles (n) of H2 gas = PV/RT

                                 = 0.99 x 0.03576/0.08205 x 294.8

                                 = 0.00146 mole

moles of Mg reacted in the sample = 0.00146 mol

mass of Mg actual present in sample = 0.00146 mol x 24.305 g/mol = 0.036 g

% m/m Mg/metal in sample = 0.036 g x 100/15.45 g = 0.233%

# error = (0.233 - 0.23) x 100/0.23 = 1.3%

2. effect on calculated % mass by the following,

a) The system had a leak - The loss of H2 would occur, therefore the final volume of H2 gas collected would be lower than actual value. Moles of H2 would be lower and thus %mass of Mg would also be lower than the actual value.

b) The levelling tank was not used - The resulting volume of H2 gas would be higher and thus moles of Mg and %mass of Mg calculated would be higher than the actual value.

c) The buret was not filled to the top before begining the experiment - The change in volume may be incorrectly read, a lower volume reading would give lower moles of H2 and moles of Mg. Therefore, lower %mass Mg would be calculated in this case.

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