Problem 8.52 When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.

Question

Problem 8.52

When a solution containing 8.00 g of NaOH in 50.0 g of water at 25.0 C is added to a solution of 8.00 g of HCl in 250.0 g of water at 25.0 C in a calorimeter, the temperature of the solution increases to 33.5 C.

Part A

Assuming that the specific heat of the solution is 4.18 J/(gC) and that the calorimeter absorbs a negligible amount of heat, calculate H in kilojoules for the reaction
NaOH(aq)+HCl(aq)NaCl(aq)+H2O(l)

Express your answer using two significant figures. NOTE: -11, -1.1x10^1, -11.0 kJ/mol are all incorrect! These are answers ive tried!

Explanation / Answer

consider the total mass of the solution

mass of solution = mass of NaOH + mass of water + mass of HCl + mass of water

mass of solution = 8 + 50 + 8 + 250

mass of solution = 316 g

now

we know that

heat = mass x specific heat x temp change

Q = m x s x dT

given

specific heat = 4.18 J / g C

so

Q = 316 x 4.18 x ( 33.5 - 25)

Q = 11227.48 J

so

heat released by the reaction = 11227.48 J

now

we know that

moles = mass / molar mass

so

moles of NaOH = 8 / 40 = 0.2

moles of HCl = 8 / 36.46 = 0.219

now

consider the given reaction

NaOH + HCl --> NaCl + H20

we can see that

moles of HCl required = moles of NaoH taken

moles of HCl required = 0.2

but

0.219 moles of HCl is present

so

HCl is in excess and NaOH is the limting reagent

now

0.2 moles of NaoH reacts with 0.2 moles of HCl to form the products

so 0.2 moles of NaOH on reaction produces 11227.48 J

now

we know that

dH = amount of heat released / moles of NaoH

dH = -11227.48 / 0.2

dH = -56137

dH = -56.137 x 1000 J /mol

dH = -56.136 kJ /mol

so

the value of dH is -56.137 kJ/mol

the sign is negative because energy is released

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