Question

What is the minimum mass of ${CaCO}_3\left(s\right)$, below which it decomposes completely, required to established equilibrium in a $6.50litre$ container for the reaction:

${CaCO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right);K_c=0.005mole/litre$

• A. $3.25g$
• B. $24.6g$
• C. $40.9g$
• D. $8.0g$

Answer 1

The correct option is A $3.25g$

For the given reaction,

$CaC{O}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+C{O}_2\left(g\right)$

Here $K_c=\left[C{O}_2\right]=0.005mo{l}^{-1}$

$\therefore$ for $6.5l$ container

moles of $C{O}_2$ required $=6.5\times 0.005$

$=0.0325mole$

$\therefore$ moles of $CaC{O}_3$ required $=0.0325mole$

mass of $CaC{O}_3$(min) $=0.0325\times 100$

$=3.25g$

Hence, the correct option is A