Explanation:
The reaction equation for the given reaction will be as follows.
![HCl + NaOH \rightarrow NaCl + H_{2}O](https://tex.z-dn.net/?f=HCl%20%2B%20NaOH%20%5Crightarrow%20NaCl%20%2B%20H_%7B2%7DO)
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 = ![\frac{mass}{molar mass of water}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bmolar%20mass%20of%20water%7D)
0.1392 moles = ![\frac{mass}{18 g/mol}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7B18%20g%2Fmol%7D)
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 = ![mS \Delta T](https://tex.z-dn.net/?f=mS%20%5CDelta%20T)
[/tex]
where, m = mass
S = specific heat capacity
= change in temperature
Putting the given values in the formula as follows.
Enthalpy of neutralization = ![mS \Delta T](https://tex.z-dn.net/?f=mS%20%5CDelta%20T)
=
= 14333.73 J
or, = 14.33 kJ
Thus, we can conclude that the enthalpy of neutralization of given reaction is 14.33 kJ.