The approximate concentration of the naoh solution you will prepare in part a of this experiment is 0.12m. calculate the number of moles and the number of grams of oxalic acid (h2c2o4 •2h20) that will be required to neutralize 25 ml of this naoh solution.

0.0015moles

0.135g

Explanation:

Given parameters:

Volume of NaOH = 25mL  = 0.025L

Molarity of NaOH = 0.12M

Unkown:

Number of moles of oxalic acid = ?

mass of oxalic acid = ?

Solution:

The reaction equation is shown below:

                              H₂C₂O₄ + 2NaOH  →  Na₂C₂O₄ + 2H₂O

   Solving:

We should work from the known to the unknown. The known here is NaOH.

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 Number of moles of NaOH = molarity of NaOH x volume = 0.12 x 0.025 = 0.003mole

##

 From the equation:

       2 moles of NaOH reacted with 1 mole of the acid

      0.03 mole of NaOH will react with = 0.0015moles of oxalic acid

##

Mass of oxalic acid = number of moles x molar mass

   Molar mass of oxalic acid = (2 x 1)  + (2 x 12) + (4 x 16) = 90g/mol

Mass of oxalic acid = 0.0015 x 90 = 0.135g of oxalic acid.

answer: -

reaction:

na + cl ⇌ nacl

balanced chemical equation:

na + 2cl ⇌ nacl2

sodium + clorine goes to form sodium chloride (table salt).

the product is nacl2.

and the reactants are na and cl.

(note: if you don’t know what ⇌ means, read below: -

⇌(double head arrow) means reversible reaction).