During decarboxylation of ethanoic acid with soda lime $(N a O H + C a O), C O_{2}$ is removed as :

C O_{2}

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C O

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N a_{2} C O_{3}

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C a C O_{3}

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Solution

The correct option is C $N a_{2} C O_{3}$
$C H_{3} - C O O N a s o d i u m e t h a n o a t e S o d a - l i m e (m i x t u r e o f N a O H a n d C a O i n 3 : 1 r a t i o) - ------------------------------- \to Δ C H_{4} M e t h a n e + N a_{2} C O_{3}$

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Similar questions

2 N a C l + C a C O_{3} \to N a_{2} C O_{3} + C a C l_{2}

C a C O_{3} Δ - \to [A] + [B]

N a C l + N H_{3} + [B] + H_{2} O \to [C] + [D]

2 [C] \to N a_{2} C O_{3} + H_{2} O + [B]

C a O

200 k g

C a {C O}_{3}

C a {C O}_{3} \to C a O + {C O}_{2}

The equilibrium constant for the reaction,

CaCO₃(s) ⇌ CaO(s) + CO₂(g), is

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