Answer:
![\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}](https://tex.z-dn.net/?f=%5Clarge%20%5Cboxed%7B%5CDelta_%7B%5Ctextbf%7Br%7D%7DH%20%3D%5Ctext%7B-4600%20J%24%5Ccdot%24%20mol%7D%5E%7B-1%7D%7D)
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}](https://tex.z-dn.net/?f=%5Crm%20%5Ctext%7B66.667%20mL%20H%24_%7B2%7D%24%7DSO_%7B4%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B3.000%20mmol%20H%24_%7B2%7D%24SO%7D_%7B4%7D%7D%7B%5Ctext%7B1%20mL%20H%24_%7B2%7D%24SO%7D_%7B4%7D%7D%20%3D%20%5Ctext%7B200.00%20mmol%20H%24_%7B2%7D%24SO%7D_%7B4%7D)
(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}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Al%28OH%29%7D_%7B3%7D%20%3D%20%20%5Ctext%7B200.00%20mmol%20of%20H%24_%7B2%7D%24SO%7D_%7B4%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B2%20mmol%20Al%28OH%29%7D_%7B3%7D%7D%7B%5Ctext%7B3%20mmol%20H%24_%7B2%7D%24SO%7D_%7B4%7D%7D%5C%5C%5C%5C%3D%20%5Ctext%7B133.33%20mmol%20Al%28OH%29%7D_%7B3%7D)
(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}](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%20Al%28OH%29%7D_%7B3%7D%20%3D%20%20%5Ctext%7B200.00%20mmol%20of%20H%24_%7B2%7D%24SO%7D_%7B4%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mL%20Al%28OH%29%7D_%7B3%7D%7D%7B%5Ctext%7B4%20mmol%20H%24_%7B2%7D%24SO%7D_%7B4%7D%7D%20%3D%20%5Ctext%7B50.000%20mL%20Al%28OH%29%7D_%7B3%7D)
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 = <u> 50.000 </u>
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}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccccl%7Dn%5CDelta%20H%20%26%20%2B%26%20mC%20%5CDelta%20T%26%20%3D%20%260%5C%5C%5Ctext%7B0.066%20667%20mol%20%7D%5Ctimes%20%5CDelta%20H%26%20%2B%20%26%20%5Ctext%7B116.667%20g%7D%20%5Ctimes%201.10%20%5Ctext%7B%20J%24%5E%7B%5Ccirc%7D%24C%24%5E%7B-1%7D%24g%24%5E%7B-1%7D%24%7D%20%5Ctimes%202.4%20%5C%2C%20%5E%7B%5Ccirc%7D%5Ctext%7BC%7D%20%26%20%3D%20%26%200%5C%5C0.066667%20%5CDelta%20H%20%5Ctext%7B%20mol%7D%20%26%20%2B%20%26%20%5Ctext%7B310%20J%7D%20%26%20%3D%20%26%200%5C%5C%26%260.066667%20%5CDelta%20H%20%5Ctext%7B%20mol%7D%20%26%20%3D%20%26%20%5Ctext%7B-310%20J%7D%20%26%20%26%20%5C%5C%5Cend%7Barray%7D%5C%5C)
![\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}}}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Bccccl%7D%26%20%26%5CDelta%20H%20%26%20%3D%20%26%20%5Cdfrac%7B%5Ctext%7B-310%20J%7D%7D%7B%5Ctext%7B0.066667%20mol%7D%7D%5C%5C%5C%5C%26%20%26%5CDelta%20H%20%26%20%3D%20%26%20%5Ctextbf%7B-4600%20kJ%2Fmol%7D%5C%5C%5Cend%7Barray%7D%5C%5C%5Clarge%20%5Cboxed%7B%5Cmathbf%7B%5CDelta_%7B%5Ctextbf%7Br%7D%7DH%7D%20%3D%5Ctextbf%7B-4600%20J%24%5Ccdot%24%20mol%7D%5E%7B%5Cmathbf%7B-1%7D%7D%7D)
This is an absurd answer, but it's what comes from your numbers.