One mole of CaCO3 is taken in one litre container which on heating decomposes according to the equation CaCO3s- CaOs-CO2gThe KP for the reaction is 1 atm at 27C- Then the maximum amount of CaO that can be produced in this reaction is

The correct option is B 2.27 g CaCo_3→ CaO+CO_2 #160; (s) #160; #160; #160; #160; #160; #160; #160; (s) #160; #160; ...

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Byju's Answer

Standard XII

Chemistry

Derivation of Kp and Kc

One mole of ...

Question

One mole of $C a C O_{3}$ is taken in one litre container which on heating decomposes according to the equation
$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
The $K_{P}$ for the reaction is 1 atm at $27^{C}$ . Then the maximum amount of CaO that can be produced in this reaction is

56g

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5.6 g

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2.27 g

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75 g

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Solution

The correct option is B 2.27 g
$C a C o_{3} \to C a O + C O_{2}$
$(s)$ $(s)$ $(g)$
Only $C O_{2}$ is an gaseous state
$K p = P c o_{2}$
$P c o_{2} = 1 a t m$ (given $K p = 1 a t m$ )
$P V = n R T$
$1 a t m \times 1 l i t r e = n \times 0.082 \times 300$
$n = \frac{1}{0.082 \times 300} = \frac{1}{24.6} = 0.0906$
$^{n} C o_{2} =^{n} C a O = 0.0906$
Amount of $C a O =$ (Molecular weight of $C a o$ ) $\times 0.0906$
$= 56 \times 0.0906$
$A m C = 2.2736 g m$

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One mole of CaCO3 is taken in one litre container which on heating decomposes according to the equationCaCO3(s)→CaO(s)+CO2(g)The KP for the reaction is 1 atm at 27C. Then the maximum amount of CaO that can be produced in this reaction is

The correct option is B 2.27 gCaCo3→CaO+CO2 (s) (s) (g)Only CO2 is an gaseous state Kp=Pco2Pco2=1 atm (given Kp=1 atm)PV=nRT1 atm×1 litre=n×0.082×300n=10.082×300=124.6=0.0906nCo2=nCaO=0.0906Amount of CaO= (Molecular weight of Cao)×0.0906=56×0.0906Am C=2.2736 gm

One mole of $C a C O_{3}$ is taken in one litre container which on heating decomposes according to the equation
$C a C O_{3} (s) \to C a O (s) + C O_{2} (g)$
The $K_{P}$ for the reaction is 1 atm at $27^{C}$ . Then the maximum amount of CaO that can be produced in this reaction is