H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
If I use 0.025 moles of NaOH to neutralize the Oxalic acid, how many moles of Oxalic acid are there?
0.025 mols NaOH = 1/2 * 0.025 mols H2C2O4
To find the number of moles of oxalic acid (H2C2O4), we can use the stoichiometry of the balanced chemical equation provided.
The balanced equation shows that 1 mole of oxalic acid reacts with 2 moles of sodium hydroxide (NaOH). Therefore, the ratio of oxalic acid to sodium hydroxide is 1:2.
Since we know that 0.025 moles of NaOH were used, we can calculate the moles of oxalic acid by multiplying the moles of NaOH by the mole ratio:
0.025 moles NaOH × (1 mole oxalic acid / 2 moles NaOH) = 0.0125 moles H2C2O4
Therefore, there are 0.0125 moles of oxalic acid present.
To find the number of moles of oxalic acid (H2C2O4) in this reaction, we can use the mole ratio between NaOH and H2C2O4. According to the balanced chemical equation:
H2C2O4(aq) + 2 NaOH(aq) → Na2C2O4(aq) + 2 H2O(l)
The mole ratio between NaOH and H2C2O4 is 2:1, which means for every 2 moles of NaOH, there is 1 mole of H2C2O4.
Given that you used 0.025 moles of NaOH, we can calculate the moles of H2C2O4 as follows:
Moles of H2C2O4 = (0.025 moles of NaOH) / (2 moles of NaOH/1 mole of H2C2O4)
Moles of H2C2O4 = 0.0125 moles
Therefore, there are 0.0125 moles of oxalic acid (H2C2O4) in the reaction.