Question

A sample of $Ca{CO}_3$ is introduced into a sealed container of volume $0.654$ litre and heated to $1000K$ until equilibrium is reached. The equilibrium is reached. The equilibrium constant for the reaction
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right)$
is $3.9\times {10}^{-2}$ atm at this temperature. Calculate the mass of $CaO$ present at equilibrium.

Answer 1

$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right)$
$K_p=p_{{CO}_2}=3.9\times {10}^{-2}atm$
Let the number of moles of ${CO}_2$ formed $=n$
$n=\frac{p_{{CO}_2}\times V}{RT}=\frac{3.9\times {10}^{-2}atm\times 0.654L}{0.082Latm/mol/K\times 1000K}=3.11\times {10}^{-4}mol$
The amount of $CaO\left(s\right)$ formed will also be $3.11\times {10}^{-4}mol$
The molecular weight of $CaO=56g/mol$
Hence, the mass of $CaO$ formed $=3.11\times {10}^{-4}mol\times 56g/mol=0.0174g$