A 0.276g sample of a CaCO3 heterogeneous mixture was placed into a gas generator

Question

A 0.276g sample of a CaCO3 heterogeneous mixture was placed into a gas generator, the total mass of which is 87.719g (containing the CaCO3 sample and the beaker, an HCl solution, and test tubes) is connected to a gas-collection apparatus. The following data were collected following the reaction.

CaCO3(s) + 2H3O(aq) --> Ca(aq) + 3H2O(l) + CO2(g)

volume of wet CO2 = 37.7mL

temperature of wet CO2 = 20.0C

pressure of wet CO2 = 770 torr

Mass of gas generator after reaction = 87.642g

a) Determone the mass and moles of CO2 evolved in the reaction. Hint: what is the mass difference of the gas generator?

b) Determine the pressure of the dry CO2.

c) calculate the volume of the dry CO2 at STP.

d) What is the calculated molar volume of the dry CO2?

e) How many moles of CaCO3 produced the moles of CO2 in the reaction?

f) Determine the mass of CaCO3 in the oringinal mixture.

g) What is the percent ( by mass) CaCO3 in the original mixture?

Thanks

Explanation / Answer

a. the mass of CO2 = 87.719-g - 87.642-g = 0.077-g moles of CO2 = .077-g / 44.0-g/mol = 0.00175 mol
b. the vapor pressure of water at 20 oC is 17.5 Torr, so the pressure of the dry CO2: 770 -17.5 =
752.5 Torr
d. PV = nRT ---> V = nRT/P 0.00175 mol x 62.396-L*Torr/mol*K x 293-K/ 752.5 Torr = 0.0425-L = 42.5-mL at 20oC and 752.5 Torr . Corrected to STP : P1V1/T1 = P2V2/T2 ---> V2 = P1V1T2/P2T1
752.5-Torr x 42.5-mL x 273-K / (760-Torr x 293-K) = 39.2-mL at STP

equation: CaCO3 + 2HCl ---> CaCl2 + CO2 + H2O
e. 0.00175 mol CaCO3 x ( 1 mol CO2 / 1 mol CaCO3) x (22,400-mL/ 1 mol CO2) = 39.2-mL CO2
f. molar ratio is 1:1 (1 mole CaCO3 produces 1 mole of CO2). Thus moles of CaCO3 is 0.00175 moles
g. 0.00175 mol CaCO3 x (100.1-g CaCO3 / 1 mol CaCO3) = 0.175-g CaCO3

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.