Laboratory Report - The Molar Volume of a Gas Data weight of magnesium in \"C te
ID: 702619 • Letter: L
Question
Laboratory Report - The Molar Volume of a Gas Data weight of magnesium in "C temperature of the water (temperature of the hydrogen) in "K 244,3 2 29 3 volume of water displaced (volume of hydrogen collected) atmospheric pressure (total gas pressure in the flask) apor pressure of water at pressure of hydrogen collected 18.6 Report (Show work for all calculations) For each run, find a. the moles of magnesium used. b. the moles of hydrogen generated. C, the volume the hydrogen would occupy at STP. d. the molar volume of hydrogen. Determine the precision of the two runs in parts per thousand by comparing the molar volumes. Determine the accuracy ofthe experiment, expressed as % error, by comparing the average expen- mental molar volume to the "true" value. Write a balanced equation for the reaction between magnesium and HCL 5) Assume the sample of magnesium used contained an inert impurity. How would this affect the Would the value be too high or too low? Explain. calculated molar volume? Assume a student made the mistakes in procedure listed below. In each case, cearly explain what a. used 0.025 grams of Mg instead of O.25 grams. b. used too much magnesium. error, if any, would be introduced. 31Explanation / Answer
Balanced equation
Mg(s) + 2 HCl(aq) --> MgCl2(aq) + H2(g)
For trial 1
Moles of Mg given = mass/molecular weight
= 0.27/24.305
= 0.0111 mol
Pressure of H2 collected = 730.7 torr x 1 atm / 760 torr
= 0.961 atm
Volume V = 0.260 L
Temperature T = 294.3 K
Moles of H2 generated = PV/RT
= 0.961 atm x 0.260 L / 0.0821 L-atm/mol-K x 294.3 K
= 0.01035 mol
Moles of Mg used = Moles of H2 generated = 0.01035 mol
At STP
Pressure P = 1 atm
Temperature T = 273 K
Volume V = nRT/P
= 0.01035 x 0.0821 x 273 / 1
= 0.2318 L = 231.8 mL
Molar volume = V/n = 0.2318 L / 0.01035 mol
= 22.4 L
Since the values are same For trial 2. The answer will also remain same
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