Question

For the decomposition of \(\mathrm{CaCO}_{3}\) at \(900^{\circ} \mathrm{C}, \mathrm{K}=1.04\). $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g) $$ Find the smallest mass of \(\mathrm{CaCO}_{3}\) needed to reach equilibrium in a \(5.00-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\).

Short answer

Answer: The smallest mass of calcium carbonate needed to reach equilibrium in a 5.00 L vessel at 900°C is 520.47 g.

Step-by-step solution

Write the balanced chemical equation and expression for K

The given balanced chemical equation is: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{O}_{2}(g) $$ Note that this equilibrium involves solids (CaCO3 and CaO) and a gas (O2). The expression for K is: $$ K = \frac{[\mathrm{O}_{2}]}{[\mathrm{CaCO}_{3}][\mathrm{CaO}]} $$ Since the concentrations of solids remain constant, we can simplify the expression to: $$ K=[\mathrm{O}_{2}] $$

Calculate the equilibrium concentration of O2

As we have K = 1.04, we can find the equilibrium concentration of O2, which will be equal to K as per the simplified expression: $$ [\mathrm{O}_{2}] = 1.04\ \mathrm{M} $$

Convert the concentration of O2 to moles

We can use the volume of the vessel and the equilibrium concentration of O2 to find the moles of O2 at equilibrium: $$ \text{moles of O}_{2} = [\mathrm{O}_{2}] \times V $$ Here, V = 5.00 L. Plugging the values, we get: $$ \text{moles of O}_{2} = 1.04\ \mathrm{M} \times 5.00\ \mathrm{L} = 5.20\ \mathrm{moles} $$

Find moles of CaCO3 needed to reach equilibrium

According to stoichiometry, 1 mole of CaCO3 will produce 1 mole of O2. Since 5.20 moles of O2 are present at equilibrium, we need the same number of moles of CaCO3 to reach equilibrium: $$ \text{moles of CaCO}_{3} = \text{moles of O}_{2} = 5.20\ \mathrm{moles} $$

Determine the mass of CaCO3 needed

Finally, we can use the molar mass of CaCO3 to convert the moles of CaCO3 needed into mass: $$ \text{mass of CaCO}_{3} = \text{moles of CaCO}_{3} \times \mathrm{Molar\ mass} $$ The molar mass of CaCO3 = 40.08 (Ca) + 12.01 (C) + 16.00 × 3 (O) = 100.09 g/mol Plugging the values, we get: $$ \text{mass of CaCO}_{3} = 5.20\ \mathrm{moles} \times 100.09\ \frac{\mathrm{g}}{\mathrm{mol}} = 520.47\ \mathrm{g} $$ So, the smallest mass of CaCO3 needed to reach equilibrium in a 5.00 L vessel at 900°C is 520.47 g.