Partial Pressure experiment, colecting gas over water. Use the information bello
ID: 506065 • Letter: P
Question
Partial Pressure experiment, colecting gas over water. Use the information bellow to calculate the mass percent of calcium carbonate using the two different methods
A. Using the Stoichiometry of reactants
What is the mass percent of calcium carbonate in the unknown sample using this data
Inital mass of sample = 0.189 g calcium carbonate/calcium chloride
initial mass of sample + reaction contatiner = 5.419 g
Final mass of sample + reaction container = 5.396 g
B. Using the ideal gas law
What is the mass percent of calcium carbonate in the unknown sample using this data
initial mass of sample = 0.189 g calcium carbonate/calcium chloride
Room temperature = 23.0 degrees C
Atmospheric pressure = 745 mmHg
Volume gas in cylinder = 11.2 to 24.1 mL
CaCO3(s) + 2HCl (aq) CaC6(aq) + Co)(q) + H2O(l)Explanation / Answer
Ans. Mass lost during reaction = 5.419 g - 5.396 g = 0.023 g
The only chemical species escaping the reaction vessel is CO2. So, the mass of CO2 produced during reaction = 0.023 g
Moles of CO2 produced = Mass/ Molar mass
= 0.023 g/ (44.0 g/ mol)
= 5.227 x 10-4 mol.
In balanced reaction, 1 mol CO2 is produced by 1 mol CaCO3. Thus, moles of CaCO3 taken is equal to moles of CO2 produced.
So,
Moles of CaCO3 = 5.227 x 10-4 mol
Theoretical Mass of CaCO3 = Moles x Molar mass
= 5.227 x 10-4 mol x (100.0872 g/ mol)
= 0.0523 g
% CaCO3 (w/ w) = (Theoretical mass of CaCO3 / Sample mass of CaCO3) x 100
= (0.0523 g / 0.189 g) x 100
= 27.67 %
#2. Increase in volume = 24.1 mL – 11.2 mL = 12.9 mL = 0.0129 L
Since, CO2 is the single chemical species in gaseous phase, increase in volume is due for formation of the gas.
So,
The gas is CO2
Volume, V = 0.0129 L
Pressure, P = 745 mm Hg = 0.98 atm ; [760 mmHg = 1.0 atm]
Temperature, T = 230C = 300.15 K
Now, 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)
Or, 0.98 atm x 0.0129 L = n (0.0821 atm L mol-1K-1) x 300.15 K
Or, n = 0.000513 mol = 5.13 x 10-4 mol
So,
Moles of CaCO3 = 5.13 x 10-4 mol
Theoretical Mass of CaCO3 = Moles x Molar mass
= 5.13 x 10-4 mol x (100.0872 g/ mol)
= 0.0513 g
% CaCO3 (w/ w) = (Theoretical mass of CaCO3 / Sample mass of CaCO3) x 100
= (0.0513 g / 0.189 g) x 100
= 27.14 %