Question

25.0 mL of a 4.21 M HCl solution at 25.2°C and 50.0 mL of a 2.50 M NaOH solution at 25.2°C are reacted together to give a final temperature of 44.1°C. The specific heat of the solution is 3.88 J/g°C and it weighs 76.6 grams. From this information, determine:

a. the limiting reactant for this reaction

b. the heat produced by the reaction (q) in units of Joules.

c. the enthalpy of the reaction (∆H) in units of kJ/mole.

Answer 1

The heat released , Q = mcdt

Where

m = mass of solution = 76.6 g

c = specific heat capacity of the solution = 3.88 J/g^oC

dt = change in temperature = final - initial

    = 44.1 - 25.2

   = 18.9 oC

Plug the values we get Q = 5617.2 J   -------- (b)

Number of moles of HCl , n = MOlarity x volume in L

                                        = 4.21 M x 0.025 L

                                         = 0.1052 mol

Number of moles of NaOH , n' = Molarity x volume in L

                                           = 2.50 M x 0.050 L

                                          = 0.125 mol

HCl + NaOH ---> NaCl + H2O

from the balanced equation,

1 mole of HCl reacts with 1 mole of NaOH

0.1052 mole of HCl reacts with 0.1052 mole of NaOH

So 0.125-0.1052 moles of NaOH left unreacted , so NaOH is the excess reactant.

Since all the mass of HCl completly reacted , so HCl is the limiting reactant   -----(a)

0.1052 moles of HCl upon produces 5617.2 J of heat

1 moles of HCl upon produces NJ of heat

N = (1x5617.2) / 0.1052

   = 53.39x10³ J /mol

   = 53.39 kJ /mol

Therefore the enthalpy of the reaction is ∆H = -53.39 kJ/mol