Question

A .276g sample of a CoCO3 heterogenous mixture was placed into a
gas generator, the total mass which is 87.719g (containing the
CaCO3 sample and the beaker, an HCl solution, and the test tubes)
is connected to a gas collection appartus. The following data were
collected following the reaction

Volume of wet CO2: 37/7 mL

Temperature of wet CO2: 20.0 degrees celcius

Pressure of wet CO2: 770 Torr

Mass of gas generator after reactionL 87.642g

Answer parts A-G and show your work

A. Determine the mass and moles of CO2 evolved in the reaction
(what is the mass difference of the gas generator)

B. Deternine the pressure of the dry
CO2

C. Calculate the volume of the dry
CO2

D. What is the calcuated molar
volume of the dry CO2

E. How many moles of CaCO3 produced
the moles of CO2

F. Determine the mass of CaCO3 in
the original mixture

G. What the percent by mass of CaCO3
in the original mixture?

Answer #1

**Solution :-**

A. Determine the mass and moles of CO2 evolved in the reaction (what is the mass difference of the gas generator)

Solution :- mass of CO2 = 87.719 g – 87.642 g = 0.077 g

Moles of CO2 = 0.077 g / 44.01 g per mol = 0.00175 mol CO2

B. Deternine the pressure of the dry CO2

Solution :- pressure of wet CO2 = 770 torr

Vapor pressure of water at 20 C = 17.5 torr

Pressure of dry CO2 = 770 torr – 17.5 torr = 752.5 torr

C. Calculate the volume of the dry CO2

Solution :- volume of CO2 = 37.7 ml = 0.0377 L

D. What is the calcuated molar volume of the dry CO2

Solution :- molar volume of dry CO2 = 0.0377 L / 0.00175 mol

= 21.5 mol / L

E. How many moles of CaCO3 produced the moles of CO2

Solution :-

Mole ratio of CaCO3 to CO2 is 1 : 1 threfore moles of CaCO3 reacted are same as moles of CO2

Moles of CaCO3 = 0.00175 mol

F. Determine the mass of CaCO3 in the original mixture

Solution :- Mass of CaCO3 = moles * molar mass

= 0.00175 mol * 100.09 g per mol

= 0.175 g

G. What the percent by mass of CaCO3 in the original mixture?

Solution :-

% of CaCO3 = (mass of CaCO3/sample mass ) *100%

=(0.175 g / 0.276 g)*100%

= 63.4 %

Constants | Periodic Table
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