Answer:
65.8 kJ
Step-by-step explanation:
M_r: 98.1
H₂SO₄ + 2LiOH ⟶ Li₂SO₄ + 2H₂O
The net ionic reaction is
H₃O⁺ + OH⁻ ⟶ 2H₂O; ΔH = -57.62 kJ·mol⁻¹
1. Calculate the moles of H₂SO₄
Moles of H₂SO₄ = 56.0 g H₂SO₄ × (98.1 g H₂SO₄/1 mol H₂SO₄)
= 0.5708 mol H₂SO₄
2. Calculate the moles of H₃O⁺
H₂SO₄ + 2H₂O ⟶ 2H₃O⁺ + SO₄²⁻
Moles of H₃O⁺ = 0.5708 mol H₂SO₄ × (2 mol H₃O⁺/1 mol H₂SO₄)
= 1.142 mol H₃O⁺
3. Calculate the energy released
q = 1.142 mol H₃O⁺ × (-57.62kJ/1 mol H₃O⁺) = -65.8 kJ
The neutralization releases 65.8 kJ.