A rigid 3.0 L vessel at 1073 K holds $CaCO X_{3} (s)$ , $CaO (s)$ and $CO X_{2} (g)$ at equilibrium, $CaCO X_{3} (s) CaO (s) + CO X_{2} (g)$ , with $K_{p} = 0.48$ atm. The vessel is then expanded isothermally to 8.0 L (solids remain present throughout). Using $R = 0.0821 L atm mo l^{- 1} K^{- 1}$ and $M (CaCO X_{3}) = 100 g mo l^{- 1}$ , calculate the additional mass of $CaCO X_{3}$ that decomposes, in g, correct to one decimal place.

Numerical answer type, answer in g, tolerance 0.1.

Answer: 2.7 g

After expansion equilibrium is restored with

p_{CO X_{2}} = K_{p} = 0.48

atm, the same as before. Moles of

CO X_{2}

then: before

= 0.48 \times 3.0/ RT

, after

= 0.48 \times 8.0/ RT

, with

RT = 0.0821 \times 1073 = 88.09

. Extra moles

= 0.48 \times 5.0/88.09 = 0.02724

mol, which equals the moles of

CaCO X_{3}

decomposed. Mass

= 0.02724 \times 100 = 2.72