C2H5OH(l) + C2H5CHOOH(l) ⇌ C2H5COOC2H5(l) + H2O(l)
7.7 mol of C2H5OH(l) and 7.32 mol of C2H5COOH(l) are reacted in a beaker, 4.88 mol of C2H5COOC2H5(l) are present at equilibrium and the total volume of liquid is exactly 2.00 L. What concentration of H2O?
The balanced equation for the reaction is:
C2H5OH(l) + C2H5CHOOH(l) ⇌ C2H5COOC2H5(l) + H2O(l)
We can use the stoichiometry of the reaction to determine the moles of water produced. From the balanced equation, we can see that for every 1 mole of C2H5OH reacted, 1 mole of water is produced. Similarly, for every 1 mole of C2H5CHOOH reacted, 1 mole of water is produced.
In this case, we know that 7.7 mol of C2H5OH and 7.32 mol of C2H5CHOOH reacted. Therefore, the number of moles of water produced is equal to the minimum number of moles of either reactant, which is 7.32 mol.
Next, we need to calculate the concentration of water in the 2.00 L solution. Concentration is defined as the amount of solute (in this case, water) divided by the volume of the solution. Therefore, the concentration of water is given by:
Concentration of water = moles of water / volume of solution
Concentration of water = 7.32 mol / 2.00 L
Concentration of water = 3.66 mol/L
Therefore, the concentration of water is 3.66 mol/L.