Answer:
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
(c). Moles of Al(OH)₃
The molar ratio is 2 mmol Al(OH)₃:3 mmol H₂SO₄
(d). Volume of 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 = <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
This is an absurd answer, but it's what comes from your numbers.