Question

A sample of $CaC{O}_3\left(s\right)$ is introduced into a sealed container of volume $0.821$ litre & heated to $1000$K until equilibrium is reached. The equilibrium constant for the reaction $CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$ is $4\times {10}^{-2}$ atm at this temperature. Calculate the mass of $CaO$ present at equilibrium.

Answer 1

Molar mass of ${CaCO}_3=100g$

Let mass of ${CaCO}_3$ initially present $=m$ ; No. of moles $=\frac{m}{100}$

Now, for the reaction ; ${CaCO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right)$

$K_p=p_{{CO}_2}$

$\Rightarrow 4\times {10}^{-2}=p_{{CO}_2}$

$\Rightarrow 4\times {10}^{-2}=\frac{nRT}{V}\left(for{CO}_2\right)$

$\Rightarrow n\left(for{CO}_2\right)=\frac{4\times {10}^{-2}\times 0.821}{0.0821\times 1000}=4\times {10}^{-4}$

So, moles of $CaO$ at equation $=4\times {10}^{-4}$

$\Rightarrow$ Mass of $CaO=4\times {10}^{-4}\times 56=0.0224g$