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

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 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

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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.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

a is the correct answer