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Question #0e2bc

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Answer 1 · 2015-05-02T21:59:40

Your reactants are acetic acid, $C H_{3} C O O H$ $CH_3COOH$ , and sodium bicarbonate, $N a H C O_{3}$ $NaHCO_3$ .

The reaction will actually take place in two steps. The first step will result in the formation of sodium acetate, $C H_{3} C O O N a$ $CH_3COONa$ , and carbonic acid, $H_{2} C O_{3}$ $H_2CO_3$ . This will be a double replacement reaction.

The carbonic acid will then undergo a decomposition reaction to produce carbond dioxide, $C O_{2}$ $CO_2$ , and water, $H_{2} O$ $H_2O$ .

$C H_{3} C O O H_{(a q)} + N a H C O_{3 (s)} \to C H_{3} C O O N a_{(a q)} + H_{2} C O_{3 (a q)}      C O_{2} + H_{2} O$ $CH_3COOH_((aq)) + NaHCO_(3(s)) -> CH_3COONa_((aq)) + underbrace(H_2CO_(3(aq)))_(color(blue)(CO_2 + H_2O))$

All in all, the reaction will look like this

$C H_{3} C O O H_{(a q)} + N a H C O_{3 (s)} \to C H_{3} C O O N a_{(a q)} + C O_{2 (g)} + H_{2} O_{(l)}$ $CH_3COOH_((aq)) + NaHCO_(3(s)) -> CH_3COONa_((aq)) + CO_(2(g)) + H_2O_((l))$

This is actually the classic vinegar - baking soda reaction.

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