Question

One mole of $CaC{O}_3$ is taken in one litre container which on heating decomposes according to the equation
$CaC{O}_3\left(s\right)\to CaO\left(s\right)+C{O}_2\left(g\right)$
The $K_P$ for the reaction is 1 atm at ${27}^C$. Then the maximum amount of CaO that can be produced in this reaction is

• A. 56g
• B. 5.6 g
• C. 2.27 g
• D. 75 g

Answer 1

Answer: (C)

2.27 g

$CaC{o}_3\to CaO+C{O}_2$

$\left(s\right)$ $\left(s\right)$ $\left(g\right)$

Only $C{O}_2$ is an gaseous state

$Kp=Pc{o}_2$

$Pc{o}_2=1atm$ (given $Kp=1atm$)

$PV=nRT$

$1atm\times 1litre=n\times 0.082\times 300$

$n=\frac{1}{0.082\times 300}=\frac{1}{24.6}=0.0906$

${}^{n}C{o}_2{=}^{n}CaO=0.0906$

Amount of $CaO=$ (Molecular weight of $Cao$)$\times 0.0906$

$=56\times 0.0906$

$AmC=2.2736gm$