Question

Question: What is the minimum mass of CaCO₃ required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant (K_c) is 0.050 for the decomposition reaction of CaCO₃ at that temperature?

Short answer

The minimum mass of CaCO₃required is 3.25 g.

Step-by-step solution

Determine amount of CO2 required

For the given reaction,

Here \({{\rm{K}}_{\rm{c}}} = \left( {{\rm{C}}{{\rm{O}}_2}} \right) = 0.005{\rm{mo}}{{\rm{l}}^{ - 1}}\)

For 6.5 L container moles of CO₂ required is

Moles of CO₂=\(6.5 \times 0.005\)

= 0.0325 mole

Determine mass of CaCO3 required

Moles of CO₂ required =0.0325 mole

Therefore, moles of CaCO₃ required = 0.0325 mole

Minimum mass of CaCO₃

\(\begin{array}{l} = \,number\,of\,moles\,of\,CaC{O_3} \times \,Molecular\,weight\,of\,CaC{O_3}\\ = 0.0325 \times 100\\ = 3.25\,gm\end{array}\)