Question

66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cup calorimeter. The initial temperature of the solutions was 22.3 C and after mixing the temperature raised to 24.7 C. If the heat capacity of the coffee cup is 1.10 J/g C, calculate the delta H reaction in J/mol Al2(SO4)3

Answer 1

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J \cdot mol}^{-1}}$

Explanation:

This is an unrealistic problem, because Al(OH)₃is highly insoluble in water.

There are two parts to this question:

A. Stoichiometry — in which we figure out the volumes, masses, and moles of products

B. Calorimetry     — in which we calculate the enthalpy of reaction.

A. Stoichiometry

1. Calculate the volume of Al(OH)₃  

(a) Balanced chemical equation.

               2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃+ 6H₂O

V/mL:                          66.667

c/mol·L⁻¹:   4.000        3 .000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H _{2} }SO_{4} \times \dfrac{\text{3.000 mmol H _{2} SO}_{4}}{\text{1 mL H _{2} SO}_{4}} = \text{200.00 mmol H _{2} SO}_{4}$

(c). Moles of Al(OH)₃

The molar ratio is 2 mmol Al(OH)₃:3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H _{2} SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H _{2} SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d). Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H _{2} SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H _{2} SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm calorimeter

 q₁   +    q₂     = 0

nΔH + mCΔT = 0

Data:

Moles  of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_i = 22.3 °C

T_f = 24.7 °C

Calculations

(a) Mass of solution

Assume the solutions have the same density as water (unrealistic).

Mass of sulfuric acid solution                =   66.667 g  

Mass of aluminium hydroxide solution =  50.000    

                                                   TOTAL =  116.667 g

(b) ΔT

ΔT = T_f - Ti = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J ^{\circ} C ^{-1} g ^{-1} } \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J \cdot mol}^{\mathbf{-1}}}$

This is an absurd answer, but it's what comes from your numbers.