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CaCO3(s)⇄CaO(s)+CO2(g) When heated strongly, solid calcium carbonate decomposes to produce solid calcium oxide and carbon dioxide gas, as represented by the equation above. A 2.0mol sample of CaCO3(s) is placed in a rigid 100.L reaction vessel from which all the air has been evacuated. The vessel is heated to 898°C at which time the pressure of CO2(g) in the vessel is constant at 1.00atm , while some CaCO3(s) remains in the vessel. (a) Calculate the number of moles of CO2(g) present in the vessel at equilibrium.

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dsdrajlin

Answer:

1.04 mol

Explanation:

CO₂ is produced in a closed 100 L vessel according to the following equation.

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

At equilibrium, the pressure of carbon dioxide remains constant at 1.00 atm.

First, we need to conver the temperature to the absolute scale (Kelvin scale) using the following expression.

K = °C + 273.15

K = 898°C + 273.15

K = 1171 K

Now, we can find the moles of carbon dioxide using the ideal gas equation.

[tex]P \times V = n \times R \times T\\n = \frac{P \times V}{R \times T} = \frac{1.00atm \times 100L}{\frac{0.0821atm.L}{mol.K} \times 1171K} = 1.04 mol[/tex]

adefioyelaoye

1.04 mole of CO₂ is present at the time of equilibrium.

CO₂ is given off in 100 L vessel according to the following equation when CaCO₃ is heated .

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

The pressure of carbon dioxide remains constant at 1.00 atm as the reaction occurred at equilibrium.

Temperature was given as 898°C

Conversion to kelvin involves

K = °C + 273.15

K = 898°C + 273.15

K = 1171 K

The moles of carbon dioxide can be calculated  using the ideal gas equation.

P× V = n × R × T

1.00 atm × 100L     = 1.04 mol of  CO₂

0.081 ₓ 1171k

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