Respuesta :
Answer:
76.0%
Explanation:
Let's consider the following reaction.
CaCO₃(s) ⇄ CaO(s) + CO₂(g)
At equilibrium, the equilibrium constant Kp is:
Kp = 1.16 = pCO₂ ⇒ pCO₂ = 1.16 atm
We can calculate the moles of CO₂ at equilibrium using the ideal gas equation.
[tex]P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{1.16atm\times 14.4 L}{(0.08206atm.L/mol.K)\times 1073K} =0.190mol[/tex]
From the balanced equation, we know that 1 mole of CO₂ is produced by 1 mole of CaCO₃. Taking into account that the molar mass of CaCO₃ is 100.09 g/mol, the mass of CaCO₃ that reacted is:
[tex]0.190molCO_{2}.\frac{1molCaCO_{3}}{1molCO_{2}} .\frac{100.09gCaCO_{3}}{1molCaCO_{3}} =19.0gCaCO_{3}[/tex]
The percentage by mass of the CaCO₃ that reacted to reach equilibrium is:
[tex]\frac{19.0g}{25.0g} \times 100\%=76.0\%[/tex]
The equilibrium constant is the ratio of the partial pressure of products to that of the reactant, the partial pressure is raised to the power of the coefficients. The percentage is 76%.
What is the ideal gas equation?
The ideal gas equation is the relation of the pressure and volume to the moles, gas constant and the temperature of the hypothetical gas.
The reaction is given as,
[tex]\rm CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)[/tex]
The equilibrium constant Kp of carbon dioxide = 1.16 atm
Substituting values in the ideal gas equation number of the moles can be estimated as,
[tex]\begin{aligned} \rm n &= \rm \dfrac{PV}{RT}\\\\&= \dfrac{1.16 \times 14.4}{0.08206\times 1073}\\\\&= 0.190 \;\rm mol\end{aligned}[/tex]
From the reaction, it can be said that 1 mole of calcium carbonate = 1 mole of carbon dioxide.
Mass of the calcium carbonate that reacted is calculated as:
[tex]\begin{aligned}0.190 \times 1 \times \dfrac{100.09}{1}= 19.0 \;\rm g \end{aligned}[/tex]
The percentage of the mass of calcium carbonate is estimated as:
[tex]\dfrac{19.0}{25.0}\times 100\%= 76\%[/tex]
Therefore, the percentage by mass of the calcium carbonate is 76%.
Learn more about the equilibrium constant here:
https://brainly.com/question/13736950
