Give the reaction of sodium carbonate with ethanoic acid :

2 C H_{3} C O O H + N a_{2} C O_{3} \to 2 C H_{3} C O O N a + C O_{2} + H_{2} O

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C H_{3} C O O H + N a_{2} C O_{3} \to C H_{3} C O O N a + N a O H + C O_{2}

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C H_{3} C O O H + N a_{2} C O_{3} \to C H_{3} C O O N a + H_{2} O + C O_{2}

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2 C H_{3} C O O H + N a_{2} C O_{3} \to 2 C H_{3} C O O N a + H_{2} C O_{3}

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Solution

The correct option is A $2 C H_{3} C O O H + N a_{2} C O_{3} \to 2 C H_{3} C O O N a + C O_{2} + H_{2} O$
Ethanoic acid + sodium carbonate --> Sodium ethanoate + sodium hydrogen carbonate
$C H_{3} C O O H$ + $N a_{2} C O_{3} \to C H_{3} C O O N a$ + $N a H C O_{3}$
OR
2 Ethanoic acid + sodium carbonate --> 2 sodium ethanoate + water + carbon dioxide
$2 C H_{3} C O O H$ + $N a_{2} C O_{3} \to 2 C H_{3} C O O N a$ + $H_{2} O$ + $C O_{2}$

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Similar questions

C H_{3} C O O N a + N a O H C a O - - \to C H_{4} + N a_{2} C O_{3}

C H_{3} C O O N a + C_{2} H_{5} O H A c i d - - \to C H_{3} C O O C_{2} H_{5} + H_{2} O

C H_{3} C O O C_{2} H_{5} + N a O H ⟶ C H_{3} C O O N a + C_{2} H_{5} O H

2 C H_{3} C O O H + 2 N a ⟶ 2 C H_{3} C O O N a + H_{2}

What are the products of the reaction, $C H_{3} C O O H + N a_{2} C O_{3} \to$ ?

{CH}_{3} COOH + {Na}_{2} {CO}_{3} \to

C H_{3} C O O N a + N a O H C a O - - \to C H_{4} + N a_{2} C O_{3}

C H_{3} C O O H + C_{2} H_{5} O H H_{2} S O_{4} - --- \to C H_{3} C O O C_{2} H_{5} + H_{2} O

2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}

C H_{3} C O O C_{2} H_{5} + N a O H \to C H_{3} C O O N a + C_{2} H_{5} O H

C H_{3} C O O N a + N a O H C a O - - \to C H_{4} + N a_{2} C O_{3}

C H_{3} C O O H + C_{2} H_{5} O H H_{2} S O_{4} - --- \to C H_{3} C O O C_{2} H_{5} + H_{2} O

2 C H_{3} C O O H + 2 N a \to 2 C H_{3} C O O N a + H_{2}

C H_{3} C O O C_{2} H_{5} + N a O H \to C H_{3} C O O N a + C_{2} H_{5} O H

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