Respuesta :
Answer:
[tex]m_{CaCO3}=10.63 g[/tex]
Explanation:
Hi, step by step:
1) In equilibrium:
[tex]P-{CO2}=0.22 atm[/tex]
Assuming ideal gas:
[tex] n_{CO2}=\frac{0.22atm*10L}{0.082\frac{atm*L}{mol*K}*385K}[/tex]
[tex] n_{CO2}=0.0697 mol[/tex]
In molarity:
[tex]M_{CO2}=\frac{0.0697mol}{10 L} =0.00697 M[/tex]
2) Equilibrium constant:
For the given chemical reaction, the constant expression is:
[tex]K=[CO_2][/tex]
Note: CaCO3 and CaO are not considered because they are in solid state
3) Concentrations in equilibrium:
Initial: 0.01 M of CaCO3 and 0.01 M of CaO
If 0.00697 M of dioxide where produced, also the same amount of CaO was produced. On the other side, 0.00697 M of CaCO3 where consumed. So:
Final: (0.01-0.00697) M of CaCO3, 0.00697 M of CO2 and (0.01+0.00697)M of CaO
4) Adding CO2
Aditional CO2: 0.24 atm:
[tex] n_{CO2}=\frac{0.24atm*10L}{0.082\frac{atm*L}{mol*K}*385K}[/tex]
[tex] n_{CO2}=0.076 mol[/tex]
The sistem will try to get back to equilibrium consuming those moles of CO2.
Following the inverse reaction, if 0.0076 M of CO2 are consumed, 0.0076 M of CaCO3 are produced:
[tex]n_{CaCO3}=((0.01-0.00697)M + 0.0076 M)*10L[/tex]
[tex]n_{CaCO3}=0.1063 mol[/tex]
The mass:
[tex]m_{CaCO3}=0.1063 mol*100g/mol[/tex]
[tex]m_{CaCO3}=10.63 g[/tex]
The CO₂ increases the mass of CaCO₃(s) and CaO in the container based
on the equilibrium constant to give approximately 10.65 g of CaCO₃(s).
Response:
- The total mass of CaCO₃(s) after equilibrium is reestablished approximately 10.65 grams
What is the equation for equilibrium constant?
The given reaction is presented as follows;
Number of moles of CaCO₃(s) = 0.100 mol
Number of moles of CaO(s) = 0.100 mol
Volume of the container = 10.0 L
Temperature to which the container is heated = 385 K
Pressure of the CO₂ after equilibrium = 0.220 atm
The chemical reaction is presented as follows;
CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Additional CO₂(g) added = 0.240 atm.
Solution:
P·V = n·R·T
[tex]n = \mathbf{\dfrac{P\cdot V}{R \cdot T}}[/tex]
Which gives;
[tex]Number \ of \ moles \ of \ CO_2, \, n = \dfrac{0.22 \times 10}{0.08205 \times 385} \approx \mathbf{0.0696}[/tex]
Therefore;
Concentration of the CO₂ = [CO₂] = 0.0696 ÷ 10 = 0.00696
The number of moles of CaCO₃(s) after the reaction = 0.1 - 0.0696 = 0.0304
The number of moles of CaO(s) after the reaction = 0.1 + 0.0696 = 0.1696
Number of moles in 0.24 atm of CO₂ is found as follows;
[tex]Number \ of \ moles \ of \ CO_2, \, n_{added} = \mathbf{\dfrac{0.24 \times 10}{0.08205 \times 385}} \approx 0.076[/tex]
The number of moles of CaCO₃ after the reaction is therefore;
[tex]n_{CaCO_3 _}equilibrium}[/tex] = 0.0304 + 0.076 = 0.1064
The molar mass of CaCO₃ = 100.0869 g/mol
Which gives;
Mass of CaCO₃ = 100.0869 g/mol × 0.1064 moles ≈ 10.65 g
- The total mass of CaCO₃(s) after equilibrium ≈ 10.65 grams
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