Question

During complete combustion of one mole of
butane, 2658 kJ of heat is released. The
thermochemical reaction for above change is
(a) 2C₄H₁₀(g) + 13O₂(g) ⟶ 8CO₂(g) + 10H₂O(l)
△cH = –2658.0 kJ mol⁻¹
(b) C₄H₁₀(g) + 13 O₂(g) ⟶ 4CO₂(g) + 5H₂O (g)
2
△cH = –1329.0 kJ mol⁻¹
(c) C₄H₁₀(g) + 13 O₂(g) ⟶ 4CO₂(g) + 5H₂O (l)
2
△cH = –2658.0 kJ mol⁻¹
(d) C₄H₁₀(g) + 13 O₂(g) ⟶ 4CO₂(g) + 5H₂O (l)
2
△cH = + 2658.0 kJ mol⁻¹

Answer 1

Answer:

(c) C₄H₁₀(g) + 13/2 O₂(g) ⟶ 4CO₂(g) + 5H₂O (l)

2

△cH = –2658.0 kJ mol⁻¹

Explanation:

Answer 2

Hello Friend

The answer of u r question is

Option.C

Thank you..!!