Name l. A student, following the procedure described in this exercise, collected
ID: 519825 • Letter: N
Question
Name l. A student, following the procedure described in this exercise, collected the following data: 0.0243 mass Mg,g 25.0 final gas volume, mL barometric pressure, torr 754 vapor pressure of H20 at 25 C 23.76 298 temperature, K *The gas collected in the eudiometer is a mixture of hydrogen and water vapor. Calculate the gas law constant. Ans. 2. What would be the volume of hydrogen gas produced by the reaction of 0.243 g ofmagnesium metal and collected at 750 torr (corrected pressure) and 298 K? Use the value of R found in question #1. Ans. 3. The results of this experiment are greatly affected by the care with which each of the steps is completed a. Describe the error that would occur if the magnesium were to slide into the HCl(aq in the beaker before the eudiometer is sealed off. b. What would be the effect on the results of a rise in the room temperature from 21.5 C when determination #1 was done, to 25.8 C, when determination #2 was made? Assume both experiments were done at constant temperature. c. Why must the surface coating of Mgo be removed prior to determining the mass of magnesium used in the experiment? 4. If the temperature and pressure remain constant, how does volume of gas sample vary with the number of moles of gas present? theExplanation / Answer
1) The reaction taking place is
Mg (s) + 2 HCl (aq) -------> MgCl2 (aq) + H2 (g) …..(1)
As per the balanced stoichiometric reaction,
1 mole Mg = 1 mole H2.
Mass of Mg taken = 0.0243 g.
Moles of Mg taken = (0.0243 g)/(24.3 g mol-1) = 0.001 mole (molar mass of Mg = 24.3 g mol-1).
Therefore, moles of H2 produced = (0.001 mole Mg)*(1 mole H2/1 mole Mg) = 0.001 mole.
Volume of the gas = 25.0 mL = (25.0 mL)*(1 L/1000 mL) = 0.025 L.
Barometric pressure = 754 torr.
Vapor pressure of water at 25C = 23.76 torr.
Therefore, pressure of dry hydrogen = (barometric pressure) – (vapor pressure of water) = (754 torr) – (23.76 torr) = 730.24 torr = (730.24 torr)*(1 atm/760 torr) = 0.961 atm.
Temperature of hydrogen gas = 25C = 298 K.
Use the ideal gas law:
P*V = n*R*T
===> R = P*V/nT = (0.961 atm)*(0.025 L)/(0.001 mole).(298 K) = 0.0806 L-atm/mol.K
The experimentally determined value of R is 0.0806 L-atm/mol.K (ans).
2) Use equation (1) and the stoichiometric balance.
Moles of Mg = (0.243 g)/(24.3 g mol-1) = 0.01 mole.
Therefore, moles of H2 produced = 0.01 mole.
The corrected pressure is 750 torr (i.e, the vapor pressure of water was subtracted) = (750 torr)*(1 atm/760 torr) = 0.987 atm.
Temperature of the gas = 298 K.
Plug in values and obtain
V = n*R*T/P = (0.01 mole)*(0.0806 L-atm/mol.K)*(298 K)/(0.987 atm) = 0.2433 L = (0.24335 L)*(1000 mL/1 L) = 243.35 mL (ans).
3a) If the Mg metal were slid into the HCl before the eudiometer was sealed off, some of the produced H2 gas will be lost to the atmosphere and hence the calculated moles of H2 will be less and hence the calculated value of R will be higher (note that R 1/n).
b) Recall that, from the ideal gas law, we have
n = P*V/RT
where n = moles of gas produced and T = absolute temperature of the gas produced.
When the temperature increases from 21.5C to 25.8C, the value of n will be low as per the above equation. Therefore, the value of R will be high (since R 1/n).
c) MgO forms a layer on the surface of Mg and prevents the metal from coming in contact with the HCl. Therefore, less Mg will be available for reaction leading to a lower yield of H2.
4) From the ideal gas law, we have
P*V = n*R*T
===> V = n*R*T/P = n*(R*T/P)
Since pressure and temperature are constant and R is also a constant, hence V can be expressed as
V = n*constant.
Therefore, the volume of the gas is directly proportional to the moles of the gas. Therefore, the higher the moles of the gas, the higher will be the volume and viceversa.
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