Respuesta :
The given 0.100 moles of CaCO₃ and CaO and the addition of 0.230 atm
of CO₂ gives a final mass of CaCO₃ of approximately 38.79 grams.
How can the mass of CaCO₃ at equilibrium be found?
Given parameters;
Number of moles of CaCO₃ = 0.100 mol
Volume of the container, V = 10.0 L
Number of moles of CaO = 0.100 mol
Temperature of the reaction, T = 385 K
Pressure of CO₂ produced, P = 0.220 atm
Pressure of CO₂ added = 0.230 atm
Required;
Mass of CaCO₃ after equilibrium is reestablished.
Solution;
Equilibrium constant for the partial pressure, [tex]\mathbf{K_p }= P_{CO_2}[/tex]
Chemical reaction; CaCO₃(s) ⇄ CaO(s) + CO₂(g)
[tex]Number \ of \ moles, \ n =\mathbf{ \dfrac{P\cdot V}{R \cdot T}}[/tex]
[tex]Moles \ of \ CO_2 \ produced = \dfrac{0.22 \ atm \times 10 \, L}{0.08206 \, L\cdot atm/\left( mol\cdot K\right) \times 385 \, K} \approx\mathbf{ 0.0696 \, moles}[/tex]
[tex]Equilibrium\ constant,\ K_c = [CO_2] \approx \dfrac{0.0696 \, moles}{10 \, L} = 0.00696 \, M[/tex]
Moles of CaCO₃ after the reaction ≈ (0.1 - 0.0696) moles = 0.304 moles
When 0.230 atm CO₂ is added, we have;
Pressure of CO₂ = 0.220 atm + 0.230 atm = 0.450 atm
Which gives;
[tex]n_{CO_2} = \dfrac{0.45 \ atm \times 10 \, L}{0.08206 \, L\cdot atm/\left( mol\cdot K\right) \times 385 \, K} \approx \mathbf{0.1532 \, moles}[/tex]
The number of moles of moles of CO₂ that react with CaO to produce
(the same number of moles of) CaCO₃, x, is found as follows;
[tex]0.00696 \ moles\approx \mathbf{\dfrac{0.1532 - x}{10} \ mol/L}[/tex]
10 × 0.00696 moles ≈ 0.1532 - x
0.1532 - 0.0696 = 0.0836 ≈ x
The final number of moles of CaCO₃ when the system reaches
equilibrium again is therefore;
New [tex]\mathbf{n_{CaCO_3}}[/tex] at equilibrium = (0.304 + 0.0836) moles = 0.3876 moles
Molar mass of CaCO₃ = 100.0869 g/mol
Final mass of CaCO₃ = 100.0869 g/mol × 0.3876 moles ≈ 38.79 grams
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