How do you balance #H_2C_2O_4 + NaOH -> Na_2C_2O_4 + H_2O#?

Answer 1

Oxalic acid + sodium hydroxide #rarr# sodium oxalate + water

Oxalic acid, #HO(O=)C-C(=O)OH#, can react with 2 equiv base in that there are 2 carboxylic acid functions:
#HO(O=)C-C(=O)OH(aq) + 2NaOH(aq) rarr {C(=O)O^-}_2Na_2^+ + 2H_2O(l)#

Is this balanced? Is there a corresponding product particle for each reactant particle?

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Answer 2

The balanced chemical equation for the reaction is:

H₂C₂O₄ + 2 NaOH → Na₂C₂O₄ + 2 H₂O

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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