unu the reaction will ker of then commence. N anment The prelab assignment will
ID: 1076298 • Letter: U
Question
unu the reaction will ker of then commence. N anment The prelab assignment will contribute 30% towards t 1 grade. The prelab assignment will be submited a minimum of 24 your hours before the beginning of the laboratory period. 1. Review the text material on Gas Laws (Ch 10: Sections 10.1-10.5 in Kotz et al. 2. The student calibrated a eudiometer by first recording the mass of the empty eudiometer followed by filling the eudiometer to the 50 mL line with deionized water and recording the mass. The mass of water was 48.894 g. Assuming "uniformity" for the calibrated scale on the Page | 3Explanation / Answer
Ans. #2. Density of water at 20.00C = 0.9982 g/ mL
Theoretical (given) volume of eudiometer = 50.0 mL
# At 20.00C, the actual/experimental/True volume of eudiometer must be equal to the volume of water filled in it.
Volume of water filled in eudiometer = Mass / Density
= 48.894 g / (0.9982 g/ mL)
= 48.9822 mL
Therefore, actual/experimental volume of eudiometer = 48.9822 mL
# We have,
Theoretical volume of eudiometer = 50.0 mL
Actual (True) of eudiometer = 48.9822 mL
Given, actual (true) gas volume = 44.4 mL
Now,
Theoretical (uncorrected) reading of 44.4 mL gas in eudiometer =
(Theoretical Eudio. Volume / Actual Eudio.) x Actual gas volume
= (50.0 mL / 48.9822 mL) x 44.4 mL
= 45.32 mL
#3. Step I: Given,
Volume of H2 gas to be collected = 44.4 mL = 0.0444 L
Temperature, T = 20.00C = 293.150C
Pressure, P = 760 mmHg = 1 atm
Using Ideal gas equation: PV = nRT - equation 1
Where, P = pressure in atm
V = volume in L
n = number of moles
R = universal gas constant= 0.0821 atm L mol-1K-1
T = absolute temperature (in K) = (0C + 273.15) K
Putting the values in above equation-
1.0 atm x 0.0444 L = n x (0.0821 atm L mol-1K-1) x 293.15 K
Or, n = 0.0444 atm L / (24.067615 atm L mol-1)
Hence, n = 0.0018448 mol
Therefore, moles of H2 to be collected = 0.0018448 mol
# Step II: Balanced reaction: Mg(s) + 2 HCl(aq) --------> MgCl2(aq) + H2(g)
According to the stoichiometry of balanced reaction, 1 mol Mg produces 1 mol H2.
So,
Required moles of Mg = 0.0018448 mol = Required moles of H2 to be produced
And,
Required mass of Mg = Required moles x Molar mass
= 0.0018448 mol x (24.305 g/ mol)
= 0.0448 g
= 44.8 mg
#4. Given, 94 mm length of Mg-ribbon of uniform dimensions has a mass of 182.0 mg.
So,
Required length of Mg-ribbon =
Required mass of Mg / (Mass per unit length of Mg-ribbon
= 44.8 mg / (182 mg metal / 94 mm)
= 23.138 mm
#5. Volume of HCl used = 1.0 mL = 0.010 L
Now,
Moles of HCl used = Molarity x Volume of solution in liters
= 4.0 M x 0.010 L
= 0.040 mol
# Amount of Mg to be reacted = 44.8 mg = 0.0018448 mol
According to the stoichiometry of balanced reaction (#3, step II), 1 mol Mg produces reacts with 2 mol HCl.
So,
Theoretical moles of HCl required = 2 x Moles of Mg reacted
= 2 x 0.0018448 mol
= 0.0036896 mol
# Since the available moles of HCl (0.04 moles) is greater than the theoretically required moles (0.0036896 mol), HCl is in excess.
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