Uhless otherWise noted, 2 pts. for each blank) Data, Trial 1 Data, Trial 2 300 A
ID: 547131 • Letter: U
Question
Uhless otherWise noted, 2 pts. for each blank) Data, Trial 1 Data, Trial 2 300 A Molarity of HCI (M) (1pt. each): Volume of HCI (mL): Mass of Mg (gram): *Volume of gas before placing in Equalization Chamber (mL) (1 pt.) (*This value NOT used = calculations below) Volume of gas after placing in Equalization Chamber (mL): Barometric Pressure (mm Hg): Temperature (C): Write the balanced overall equation for the reaction of magnesium and HCl to form hydrogen gas and the magnesium salt (2 pts): Mole of Mg reacted Mole of H2 formed (based on the stoichiometric equation for the reaction) Vapor Pressure of water from curve in mmHg: Pressure of H: from Dalton's Law in mmu Pressure of H2 in atm Volume of H2 in liter (Using volume measured in Equalization Chamber) Temperature in Kelvin Calculation of R (6 pts. each) Percent error Average RExplanation / Answer
Balanced overall equation: Mg(s) + 2HCl(aq) ----> MgCl2 (aq) + H2 (g)
For trial 1 :
Mole of Mg = 0.041 gm / 24.3056 gm/mol = 1.6868*10-3 mol
1 mol of Mg gives 1 mol of H2
Mole of H2 formed = 1.6868*10-3 mol
At 21.4 Vp of water = 18.955 mm Hg
pressure of H2 = 738.9 - 18.955 = 719.945 mm Hg = 0.9472 atm
volume of H2 in L = 42.5/1000 = 0.0425L
Temperature in K = 21.4+273.15 = 294.55 K
R = PV / nT = (0.9472 atm * 0.0425L) / (1.6868*10-3 mol*294.55 K)
Rcalculated = 0.081022 atm L mol-1 K-1
Ractual = 0.0821 atm L mol-1 K-1
% error in R = [ |Ractual - Rcalc| / Ractual ] *100
% error in R = [ |0.0821-0.081022| / 0.0821]*100 = 1.313 %
For trial 2 :
Mole of Mg = 0.044 gm / 24.3056 gm/mol = 1.8102*10-3 mol
1 mol of Mg gives 1 mol of H2
Mole of H2 formed = 1.8102*10-3 mol
At 21.4 Vp of water = 18.955 mm Hg
pressure of H2 = 738.9 - 18.955 = 719.945 mm Hg = 0.9472 atm
volume of H2 in L = 19/1000 = 0.019L
Temperature in K = 21.4+273.15 = 294.55 K
R = PV / nT = (0.9472 atm * 0.019L) / (1.8102*10-3 mol*294.55 K)
Rcalculated = 0.03375 atm L mol-1 K-1
Ractual = 0.0821 atm L mol-1 K-1
% error in R = [ |0.0821-0.03375| / 0.0821]*100 = 58.88 %
Note 2 : the error is large, maybe there are some major sources of error in calibration of volume.
Mass of Mg = 0.041 gm (consider for trial 1)
Mol of Mg = 0.041 gm / 24.3056 gm/mol = 1.6868*10-3 mol
According to the reaction stoichiometry,
1 mol of Mg reacts with 2 mol of HCl so mol of HCl will be 2 times that of Mg
i.e) 2*1.6868*10-3 = 3.3736*10-3 mol
This shows that Mg is indeed the limiting reactant.
The pressure outside the tube is greater than the inside atmospheric pressure.
This you will know by comparing the volumes as it was asked in the question, the liquid
level is higher than level when placed inside the chamber.
Pressure measurement most of the times will be inaccurate, to get close values of R
sometimes corrected pressures are used.
The following experimental changes might improve the accuracy in the determined R value
(i) Adding exact volumeof HCl into eudiometer tube (ii) Recording the temperature after
perfect equilibrium is attained (iii) weighing the Mg i.e) even considering at least 4 significant figures
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