Respuesta :
Answer:
The mass percentage of calcium carbonated reacted is 2.5%.
Explanation:
The reaction is:
[tex]CaCO_{3}(s)--->CaO(s)+CO_{2}(g)[/tex]
Thus the Kp of the equilibrium will be:
Kp = partial pressure of carbon dioxide [as the other are solid]
Moles of calcium carbonate initially present = [tex]\frac{mass}{molarmass}=\frac{20}{100}=0.2[/tex]
Let us apply ICE table to the equilibrium given:
[tex]CaCO_{3}(s)--->CaO(s)+CO_{2}(g)[/tex]
Initial 0.2 0 0
Change -x +x +x
Equilibrium 0.2-x x x
Kp = partial pressure of carbon dioxide
Kp = Kc(RT)ⁿ
where n = difference in the number of moles of gaseous products and reactants
for given reaction n = 1
R = gas constant = 8.314 J /mol K
T = temperature = 800 ⁰C = 1073 K
Putting values
Kc =[tex]\frac{Kp}{RT}=\frac{1.16}{8.314X1073}=1.3X10^{-4}[/tex]
Kc = [tex]\frac{[CO_{2}][CaO]}{[CaCO_{3}]}= \frac{x^{2} }{(0.2-x)}=1.3X10^{-4}[/tex]
[tex]1.3X10^{-4}(0.2-x)=x^{2}[/tex]
[tex]x^{2} = 0.26X10^{-4}-1.3X10^{-4}x[/tex]
On calculating
x = 0.005
where x = the moles of calcium carbonate dissociated or reacted.
Percentage of the moles or mass reacted = [tex]\frac{molesreacted X100}{initialmoles}=\frac{0.005X100}{0.2}=2.5[/tex]%
Kp is the equilibrium constant of the reaction in accordance with the partial pressure of the gas. The mass percentage of calcium carbonate is 2.5%.
What is the mass percentage?
The mass percentage is the ratio of the moles reacted to that of the moles present initially multiplied by 100.
The reaction of calcium carbonate can be shown as,
[tex]\rm CaCO_{3} \rightarrow CaO+CO_{2}[/tex]
The formula for the equilibrium constant is given as,
[tex]\rm Kp = Kc(RT)^{n}[/tex]
Here,
n = 1
Gas constant (R) = 8.314 J /mol K
Temperature (T) = 1073 K
Kp = 1.16
Substituting values in the above equation:
[tex]\begin{aligned}\rm Kc &= \rm \dfrac{Kp}{RT}\\\\&= \dfrac{1.16}{8.314\times 1073}\\\\&= 1.3 \times 10^{-4}\end{aligned}[/tex]
From the ICE table of the reaction the value of moles of calcium carbonate (x) is calculated as:
[tex]\rm Kc = \dfrac{[CO_{2}][CaO]}{[CaCO_{2}]}= \dfrac{x^{2}}{(0.2-x)}= 1.3\times 10^{-4}[/tex]
Solving further,
[tex]\begin{aligned} \rm 1.3\times 10^{-4}(0.2-x) &= \rm x^{2}\\\\\rm x^{2} &= \rm 0.26\times 10^{-4} - 1.3 \times 10^{-4}x\\\\\rm x &= 0.005\end{aligned}[/tex]
The mass percentage of the calcium carbonate is given by:
[tex]\begin{aligned}\rm mass\;\% &= \dfrac{\text{moles reacted}}{\text{initial moles}}\times 100\%\\\\&= \dfrac{0.005}{0.2}\times 100\%\\\\&= 2.5\%\end{aligned}[/tex]
Therefore, 2.5% is the mass percentage.
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