How do you balance ${(N H_{4})}_{2} C O_{3} + N a O H \to N a_{2} C O_{3} + N H_{3} + H_{2} O$ $(NH_4)_2CO_3 + NaOH -> Na_2CO_3 + NH_3 + H_2O$ ?

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How do you balance ${(N H_{4})}_{2} C O_{3} + N a O H \to N a_{2} C O_{3} + N H_{3} + H_{2} O$ $(NH_4)_2CO_3 + NaOH -> Na_2CO_3 + NH_3 + H_2O$ ?

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Answer 1 · 2016-06-19T05:45:28

${(N H_{4})}_{2} C O_{3} (a q) + 2 N a O H \to N a_{2} C O_{3} (a q) + 2 N H_{3} (a q) + 2 H_{2} O (l)$ $(NH_4)_2CO_3(aq) + 2NaOH rarr Na_2CO_3(aq) + 2NH_3(aq) + 2H_2O(l)$

Explanation:

The net ionic equation is simply an acid base reaction:

$N H_{4}^{+} + H O^{-} \to N H_{3} (a q) + H_{2} O (l)$ $NH_4^+ + HO^(-) rarr NH_3(aq) + H_2O(l)$

Given that a mineral acid will protonate a carbonate to give water and carbon dioxide gas, what would the reaction if sodium carbonate were treated with 2 equiv hydrochloric acid?

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How do you balance ${(N H_{4})}_{2} C O_{3} + N a O H \to N a_{2} C O_{3} + N H_{3} + H_{2} O$ $(NH_4)_2CO_3 + NaOH -> Na_2CO_3 + NH_3 + H_2O$ ?

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Explanation:

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How do you balance (NH4)2CO3+NaOH→Na2CO3+NH3+H2O(NH_4)_2CO_3 + NaOH -> Na_2CO_3 + NH_3 + H_2O?

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Explanation:

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How do you balance ${(N H_{4})}_{2} C O_{3} + N a O H \to N a_{2} C O_{3} + N H_{3} + H_{2} O$ $(NH_4)_2CO_3 + NaOH -> Na_2CO_3 + NH_3 + H_2O$ ?