How do you balance $H_{2} C_{2} O_{4} + N a O H \to N a_{2} C_{2} O_{4} + H_{2} O$ $H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O$ ?

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How do you balance $H_{2} C_{2} O_{4} + N a O H \to N a_{2} C_{2} O_{4} + H_{2} O$ $H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O$ ?

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Answer 1 · 2016-01-15T18:01:12

Oxalic acid + sodium hydroxide $\to$ $rarr$ sodium oxalate + water

Explanation:

Oxalic acid, $H O (O =) C - C (= O) O H$ $HO(O=)C-C(=O)OH$ , can react with 2 equiv base in that there are 2 carboxylic acid functions:

$H O (O =) C - C (= O) O H (a q) + 2 N a O H (a q) \to {C (= O) O^{-}}_{2}^{-} N a_{2}^{+} + 2 H_{2} O (l)$ $HO(O=)C-C(=O)OH(aq) + 2NaOH(aq) rarr {C(=O)O^-}_2Na_2^+ + 2H_2O(l)$

Is this balanced? For every reactant particle is there a corresponding product particle?

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How do you balance $H_{2} C_{2} O_{4} + N a O H \to N a_{2} C_{2} O_{4} + H_{2} O$ $H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O$ ?

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How do you balance H2C2O4+NaOH→Na2C2O4+H2OH_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O?

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How do you balance $H_{2} C_{2} O_{4} + N a O H \to N a_{2} C_{2} O_{4} + H_{2} O$ $H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O$ ?