Question

Consider the reaction: 2HCl(aq) + Ba(OH)2(aq) ! BaCl2(aq) + 2 H2O(l) deltaH = "118 kJ

A) Calculate the heat produced when 700.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.500 M Ba(OH)2. Note: One of the solutions is the limiting reagent.

B) Calculate the final temperature if both solutions were initially at 25.0° C, and they were mixed together in a calorimeter with a heat capacity of 180.3 J/C° Note: The density specific heat capacity of the solution are the same as that of water: D = 1.00 g/cm3 and C = 4.184 J /g•°C

Answer 1

HCl moles = 0.500 x 700 /1000 =0.35

Ba(OH)2 moles= 0.5 x300 / 1000= 0.15

2HCl(aq) + Ba(OH)2(aq) -----------------> BaCl2(aq) + 2 H2O(l) deltaH = "118 kJ

2mole ------------> 1 mole Ba(OH)2 --------------------> 118 kJ

0.35 --------------->   0.15 ---------------------------------------> ?

here limiting reagent Ba(OH)2 . the product depends on moles of Ba(OH)2.

1 mole Ba(OH)2 ---------------------> 118 kJ

0.15 mole --------------> ??

heat produced = 118 x 0.15

                        = 17. 7 kJ

2) reaction between acid +base gives H2O . so the mass of H2O is needed

for 1 mole Ba(OH)2 -----------------> 2 mole H2O

   0.15 mole ------------------------> ?

moles of water = 2 x 0.15

                         = 0.3

weight of water = 0.3 x 18 = 5.4 g

Q = m Cp (T2-T1)

17.7 = 5.4 x 4.184 x (T2 - 25)

T2 = 25.78 C^o

final temperature = T2 = 25.78 C^o