When $\mathbf{\text{0.100mol}}$of${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\left(}\mathit{s}\mathbf{\right)}$ and$\mathbf{\text{0.100mol}}$ mol of$\mathbf{\text{CaO(s)}}$ are placed in an evacuated sealed$\mathbf{\text{10.0-L}}$ container and heated to$\mathbf{\text{385K}}$ ,${\mathbf{\text{P}}}_{{\mathbf{\text{CO}}}_{\mathbf{\text{2}}}}\mathbf{\text{=0.220atm}}$ after equilibrium is established:

${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}\mathbf{\text{(s)}}\mathbf{\rightleftharpoons }{\mathbf{\text{CaO(s)+CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$

An additional$\mathbf{\text{0.300atm}}$ of ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}\mathbf{\text{(g)}}$is pumped in. What is the total mass (inrole="math" localid="1656942360456" $\mathbf{g}$ ) of${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}$ after equilibrium is re-established?

Chemical equilibrium is the state of a system in which the concentration of the reactant and the concentration of the products do not change over time and the system's attributes do not change.

The reaction given is –

${\text{CaCO}}_3\text{(S)}\rightleftharpoons {\text{CaO(s)+CO}}_2\text{(g)}$

First write the expression for the equilibrium constant of the reaction in terms of partial pressures of the reactants. It should only include gaseous reaction components.

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}\frac{{\text{P}}_{\text{products}}}{{\text{P}}_{\text{reactants}}}\\ {\text{K}}_{\text{p}}{\text{=P}}_{{\text{CO}}_{\text{2}}}\end{array}$

Therefore,${\mathrm{K}}_{\mathrm{P}}$is just equal to the partial pressure ${\text{CO}}_2$.

At the first equilibrium, the moles of ${\text{CO}}_2$ can be solved using the ideal gas law equation.

$\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\\ \text{=}\frac{\text{(0.220atm)(10.0L)}}{\left(\text{0.0821}\frac{\text{atm×L}}{\text{mol×K}}\right)\text{(385K)}}\\ {\text{n=0.0696molCO}}_{\text{2}}\end{array}$

Using stoichiometry, we can solve for the amount of moles used to produce ${\text{0.0696molCO}}_{\text{2}}$.

$\begin{array}{c}{\text{molCaCO}}_{\text{3}}{\text{=0.0696molCO}}_{\text{2}}\text{×}\frac{{\text{1molCaCO}}_{\text{3}}}{{\text{1molCO}}_{\text{2}}}\\ {\text{=0.0696molCaCO}}_{\text{3}}\end{array}$

Subtract it from the initial mol of ${\text{CaCO}}_{\text{3}}$present.

$\begin{array}{c}{\text{molCaCO}}_{\text{3}}{\text{left=(0.100-0.0696)molCaCO}}_{\text{3}}\\ {\text{=0.0304molCaCO}}_{\text{3}}\end{array}$

The reaction table is –

Substitute the equilibrium formula from our reaction table to the expression for the equilibrium constant in terms of partial pressures.

$\begin{array}{c}{\text{K}}_{\text{p}}{\text{=P}}_{{\text{CO}}_{\text{2}}}\\ \text{0.220=0.520+x}\\ \text{x=-0.300atm}\end{array}$

The partial pressure of ${\mathrm{CO}}_2$ decreased by $\text{0.300atm}$. Therefore, ${\text{CaCO}}_{\text{3}}$ also increased with the same amount in terms of moles (given that their mole ratio is $1:1$ from the previous calculation). Solve for the number of moles of ${\text{CO}}_2$using the ideal gas law equation.

$\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\\ \text{=}\frac{\text{(0.300atm)(10.0L)}}{\left(\text{0.0821}\frac{\text{atm×L}}{\text{mol×K}}\right)\text{(385K)}}\\ {\text{n=0.0949molCO}}_{\text{2}}\end{array}$

Since the mole ratio of ${\text{CO}}_2$and ${\text{CaCO}}_{\text{3}}$ is $1:1$, then mol ${\text{CO}}_2{\text{=molCaCO}}_{\text{3}}$. Add it from the previous mol of ${\text{CaCO}}_{\text{3}}$calculated.

data-custom-editor="chemistry" $\begin{array}{c}{\text{molCaCO}}_{\text{3}}{\text{left=(0.0304+0.0949)molCaCO}}_{\text{3}}\\ {\text{=0.0125molCaCO}}_{\text{3}}\end{array}$

Solve for the mass of data-custom-editor="chemistry" ${\text{CaCO}}_{\text{3}}$using dimensional analysis.

data-custom-editor="chemistry" $\begin{array}{c}{\text{massCaCO}}_{\text{3}}{\text{left=0.0125molCaCO}}_{\text{3}}\text{×}\frac{{\text{100.09gCaCO}}_{\text{3}}}{{\text{1molCaCO}}_{\text{3}}}\\ {\text{=12.5gCaCO}}_{\text{3}}\end{array}$

Therefore, the value of mass is obtained as${\text{12.5gCaCO}}_{\text{3}}$ .