A .276g sample of a CoCO3 heterogenous mixture was placed into a gas generator,

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?

Explanation / Answer

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 %

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