An unkown vol of 2Mol of NaOH reacted with 3g of ethanoic acid .....get the volume of NaOH
To find the volume of NaOH that reacted with the given amount of ethanoic acid, we need to use the concept of stoichiometry to determine the balanced equation for the reaction between NaOH and ethanoic acid.
First, let's write the balanced chemical equation for the reaction:
C2H4O2 + NaOH → NaC2H3O2 + H2O
From the balanced equation, we can see that the stoichiometric ratio between ethanoic acid and NaOH is 1:1. This means that one mole of ethanoic acid reacts with one mole of NaOH.
Given that 3 grams of ethanoic acid reacted, we need to convert this into moles. The molar mass of ethanoic acid (C2H4O2) is approximately 60.05 g/mol.
Number of moles of ethanoic acid = Mass / Molar mass
= 3 g / 60.05 g/mol
= 0.0499 mol (rounded to four decimal places)
Since the stoichiometric ratio between ethanoic acid and NaOH is 1:1, we can conclude that 0.0499 moles of NaOH reacted.
Finally, to determine the volume of NaOH, we need to know the concentration of the NaOH solution. You mentioned that the unknown volume of a 2 M solution of NaOH reacted.
Using the definition of molarity (M):
Molarity (M) = Moles / Volume (in liters)
From the given molarity of 2 M, we can rearrange the equation to solve for volume:
Volume (in liters) = Moles / Molarity
= 0.0499 mol / 2 M
= 0.02495 L (rounded to five decimal places)
Therefore, the volume of NaOH that reacted is approximately 0.02495 liters or 24.95 milliliters.