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Molarity

CaC O 3 s ⇌ ...

Question

$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$ ; $K_{p} = 1.64 a t m$ at $1000 K$
$50 g$ of $C a C O_{3}$ in a $10$ litre closed vessel is heated to $1000 K$ . Percentage of $C a C O_{3}$ that remains unreacted at equilibrium is :
(Given $R = 0.082 L a t m K^{- 1} {m o l}^{- 1}$ )

50

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20

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40

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60

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Solution

The correct option is C $60$
Number of moles of $C a C O_{3}$ present initially $= \frac{50}{100} = 0.5 m o l e s$

$K_{p}$ of the given reaction can be written as:

$K_{p} = P_{C O_{2}} = 1.64 a t m$

Number of moles of $C O_{2} = \frac{P_{C O_{2}} V}{R T} = \frac{1.64 \times 10}{0.082 \times 1000} = 0.2$

Number of moles of $C a C O_{3}$ reacted $= 0.2 m o l e s$

Number of moles of $C a C O_{3}$ unreacted $= (0.5 - 0.2) = 0.3 m o l e s$

Percentage of $C a C O_{3}$ unreacted $= \frac{0.3}{0.5} \times 100 = 60 %$

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CaCO3(s)⇌CaO(s)+CO2(g); Kp=1.64 atm at 1000K50g of CaCO3 in a 10 litre closed vessel is heated to 1000K. Percentage of CaCO3 that remains unreacted at equilibrium is :(Given R=0.082 L atm K−1mol−1)

$C a C O_{3 (s)} ⇌ C a O_{(s)} + C O_{2 (g)}$ ; $K_{p} = 1.64 a t m$ at $1000 K$
$50 g$ of $C a C O_{3}$ in a $10$ litre closed vessel is heated to $1000 K$ . Percentage of $C a C O_{3}$ that remains unreacted at equilibrium is :
(Given $R = 0.082 L a t m K^{- 1} {m o l}^{- 1}$ )