Calculate and compare the value for the molar volume at 21.5 °C obtained in part
ID: 484262 • Letter: C
Question
Calculate and compare the value for the molar volume at 21.5 °C obtained in part 1 with the value for the molar volume at 21.5 °C obtained in part 2. The molar mass of Zn is 65.38 g/mol. To receive full points for this question, list every step of your calculation.
Please use data below.
There are 6M HCl and all of them were 10.00mL volumes(same) 0.250g of zinc, and 0.500g of zinc were used in each of the trials. (balanced)
Zn(s) + 2HCl (aq) H2 (g) + ZnCl2 (aq)
Trial 1
Trial 2
Trial 3
a
Mass of zinc used (g)
.250
.500
1.00
b
Volume of 6.00 M HCl used (mL)
10.0
10.0
20.0
c
Molar mass of zinc (g/mol)
65.39
65.39
65.39
d
Moles of zinc reactant
.00382
.00765
.0153
e
Moles of HCl reactant (Calculate from Mand V)
0.06
0.06
.12
f
Identity of Limiting Reactant
Zinc
Zinc
Zinc
g
Moles of hydrogen produced calculated using stoichiometry from limiting reactant (mol)
.00382
.00765
.0153
h
Volume of gas produced at initial room temperature of 21.5C (mL)
92.2
184.5
Glass broke
But shouldve been approximately = 369 mL
i
Calculated Molar Volume of the ideal hydrogen gas at room temperature (Volume/moles), expressed in units ofL/mol (use unrounded value of moles H2 for calculation) Calculated from your experimental data
24.14
24.12
24.12
j
Under what conditions of temperature and pressure is this molar volume value valid? Include units.
0.00 degrees Celsius and 1 atm of pressure
0.00 degrees Celsius and 1 atm of pressure
0.00 degrees Celsius and 1 atm of pressure
Zn(s) + 2HCl (aq) H2 (g) + ZnCl2 (aq)
Explanation / Answer
Since - PV = nRT
Molar volume, (V/n) = RT / P
Trail:1
Molar volume from the given data = (0.0922 L / 0.00382 mol) = 24.14 L/mol
Trail:2
Molar volume from the given data = (0.1845 L / 0.00765 mol) = 24.14 L/mol
So in both conditions molar volume is equal
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