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

Question

During decarboxylation of ethanoic acid with soda lime -NaOH-CaO-  CO-2 is removed as -COCO-2Na-2CO-3CaCO-3

Toppr Admin · Accepted Answer

CH3-x2212-COONasodiumethanoateSoda-x2212-lime-mixtureofNaOHandCaOin3-1ratio-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2212-x2192-x394-CH4Methane-Na2CO3

During decarboxylation of ethanoic acid with soda lime $(N a O H + C a O), C O_{2}$ is removed as :
$C O$
$C O_{2}$
$N a_{2} C O_{3}$
$C a C O_{3}$

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

During decarboxylation of ethanoic acid with soda lime (NaOH+CaO), CO2 is removed as :COCO2Na2CO3CaCO3

CH3−COONasodiumethanoateSoda−lime(mixtureofNaOHandCaOin3:1ratio)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−→ΔCH4Methane+Na2CO3

During decarboxylation of ethanoic acid with soda lime $(N a O H + C a O), C O_{2}$ is removed as :
$C O$
$C O_{2}$
$N a_{2} C O_{3}$
$C a C O_{3}$

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