Question

The following information was obtained when carrying out an experiment to determine the Enthalpy (ΔH) of neutralization reaction between HCl and NaOH. 100 mL of 2.00 M HCl solution was added to 95.00 mL of 2.00 M solution of NaOH. The final temperature reached was 35.40 °C and the initial temperature at mixing was 22.15 °C. The density of the reaction mixture was 1.04 g/mL and its specific heat capacity was 3.89 J/g-°C. Calculate the weight of the reaction mixture?

Answer 1

total volume of solution = (100 + 95.00) = 195 ml

we know,

density = mass / volume

thus

weight of the reaction mixture = 195 ml * 1.04 g / ml = 202.8 g

100 mL of 2.00 M HCl solution = 0.100 L* 2.00 mole / L = 0.200 mole HCl.

95.00 mL of 2.00 M solution of NaOH = 0.095 L * 2.00 mole / L = 0.19 mole.

NaOH + HCl ............> NaCl + H2O

thus

NaOH is the limiting reactant.

Heat = m * s * dT = 202.8 g * 3.89 J / g-oC * (35.40 - 22.15) oC = 10452.8 J

Enthalpy (ΔH) of neutralization = - 10452.8 J / 0.19 mole = - 55014.8 J / mole = - 55.0 KJ / mole.