How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?

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How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?

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Answer 1 · 2015-11-03T15:22:40

$N a_{2} C O_{3} (s) + 2 H C l (a q) \to 2 N a C l (a q) + H_{2} O (l) + C O_{2} (g)$ $Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)$

Explanation:

$N a_{2} C O_{3} (s) + 2 H C l (a q) \to 2 N a C l (a q) + H_{2} O (l) + C O_{2} (g)$ $Na_2CO_3(s) + color(red)(2)HCl(aq) -> color(red)(2)NaCl(aq) + H_2O(l) + CO_2(g)$

The reasoning that you should follow in order to balance this reaction easily is:

1- Look at the $N a$ $Na$ . You have $2$ $2$ in $N a_{2} C O_{3}$ $Na_2CO_3$ and $1$ $1$ in $N a C l$ $NaCl$ then you should multiply $N a C l$ $NaCl$ by $2$ $color(red)(2)$ .

2- When multiplying $N a C l$ $NaCl$ by 2, we will have to $C l$ $Cl$ in the products side, therefore, we should multiply $H C l$ $HCl$ by $2$ $color(red)(2)$ .

3- Look at the carbon atom $C$ $C$ there is one atom at each side. No need for any action.

4- Look at the hydrogen atom $H$ $H$ there is 2 atoms at each side. No need for any action.

5- Look at the oxygen atom $O$ $O$ there is 3 atoms in $N a_{2} C O_{3}$ $Na_2CO_3$ and there is 1 atom in $H_{2} O$ $H_2O$ and 2 atoms in $C O_{2}$ $CO_2$ which makes it a total of 3 atoms. No need for any action.

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How would you balance Na2CO3(s) + HCl(aq) = NaCl(aq) + H2O(l) + CO2(g)?

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