Explanation:
The reaction equation for the given reaction will be as follows.
Each mole of both HCl and NaOH gives one mole of water.
Also, it is given that 1 liter of NaOH and HCl solution contains 1.6 mol of NaOH (HCl).
It is known that 1 = 0.001 liter. So, 87 = 0.087 liter.
Hence, number of moles of water obtained from the given reaction are as follows.
0.087 liter × 1.6 mol = 0.1392 moles
No. of moles =
0.1392 moles =
mass = 2.5056 g
Now, volume of water present before the reaction is liter = 0.174 liter or 0.174 Kg (as density is 1 ) or 174 g (as 1 kg = 1000 g).
Therefore, total weight of water present = 2.5056 g + 174 g = 176.5056 g
Formula to calculate enthalpy of neutralization is as follows.
Enthalpy of neutralization =
[/tex]
where, m = mass
S = specific heat capacity
= change in temperature
Putting the given values in the formula as follows.
Enthalpy of neutralization =
=
= 14333.73 J
or, = 14.33 kJ
Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.