Respuesta :
Explanation:
The given data is as follows.
Densities of both the reactant solutions = 1.07 g/mL.
Specific heat = 4.18 [tex]J/g^{o}C[/tex]
Volume of HCl = 32.8 ml
Volume of NaOH = 66.3 ml
The reaction will be as follows.
[tex]HCl + NaOH \rightarrow NaCl + H_{2}O[/tex]
Since, density is mass divided by volume. Hence, calculate the mass of HCl as follows.
Mass of HCl = Density × Volume of HCl
= [tex]1.07 \times 32.8 ml[/tex]
= 35.096 g
Similarly, mass of NaOH will be calculated as follows.
Mass of NaOH = Density × Volume of NaOH
= [tex]1.07 g/ml \times 66.3 ml[/tex]
= 70.941 g
Therefore, total mass will be as follows.
Total mass = Mass of HCl + Mass of NaOH
= 35.096 g + 70.941 g
= 106.037 g
Change in temperature will be calculated as follows.
[tex]\Delta T[/tex] = [tex](28.2 - 25)^{o}C[/tex]
= [tex]3.2^{o}C[/tex]
Therefore, calculate the heat of reaction as follows.
q = [tex]mC \Delta T[/tex]
= [tex]106.037 g \times 4.18 J/g^{o}C \times 3.2^{o}C[/tex]
= 1418.35 J
or, = 1.418 kJ (as 1 kJ = 1000 J)
No. of moles of HCl = Molarity × Volume
= 0.5 M × 32.8 ml
= 16.4 mol
No. of moles of NaOH = Molarity × Volume
= 0.5 M × 66.3 ml
= 33.15 mol
So, HCl is the limiting reagent and heat of reaction produces by per mole of NaCl will be calculated as follows.
Heat released for 1 mole of NaCl = [tex]\frac{1.418 kJ}{16.4 mol}[/tex]
= 0.0864 kJ/mol
Thus, we can conclude that the heat of reaction per mole of NaCl is 0.0864 kJ/mol.
