Complete combustion of 75.0 g acetylene, C2H2, released 3750 kJ of heat. What is the ΔH
(heat of combustion)of acetylene? (Hint: you’ll need to write a balanced combustion reaction)
I wrote the combustion reaction as: 2 C2H2 5 O2 --> 4 CO2 2 H2O
The answer is -1302 kJ/mol but I can't seem to get it right, any help explaining how to get this answer would be great!
Complete combustion of 75.0 g acetylene, C2H2, released 3750 kJ of heat. What is the ΔH of acetylene?
2016-06-13 6:39 am
回答 (2)
2016-06-18 12:22 pm
✔ 最佳答案
moles C2H2 = mass / molar mass = 75.0 g / 26.036 g/mol = 2.8806 mol All of these properties are always for 1 mole of the specified compound
So when it wants heat of combustion of C2H2, it means for the reaction of 1 mole of C2H2
If it were to ask for the heat of formation of CO2, it means for 1 mole of CO2 to be produced.
etc
Now, we know 2.8806 moles C2H2 releases 3750 kJ (releases = -ve)
So ΔH of 2.8806 mol C2H2 = -3750 kJ
ΔH 1 mol C2H2 = -3750 kJ x (1 mol C2H2 / 2.8806 mol C2H2)
= -1302 kJ
C2H2 2.5 O2 --> 2 CO2 H2O, ΔH = -1302 kJ/mol
2016-06-13 6:46 am
C₂H₂(g) + (5/2)O₂(g) → 2CO₂(g) + H₂O(l) ...... ΔHc[C₂H₂(g)]
Molar mass of C₂H₂ = (12.01×2 + 1.01×2) g/mol = 26.04 g/mol
No. of moles of C₂H₂ burned = (75.0 g) /(26.04 g/mol) = 2.88 mol
ΔHc[C₂H₂(g)] = (-3750 kJ) / (2.88 mol) = -1302 kJ/mol
Molar mass of C₂H₂ = (12.01×2 + 1.01×2) g/mol = 26.04 g/mol
No. of moles of C₂H₂ burned = (75.0 g) /(26.04 g/mol) = 2.88 mol
ΔHc[C₂H₂(g)] = (-3750 kJ) / (2.88 mol) = -1302 kJ/mol
收錄日期: 2021-04-18 14:59:25
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160612223904AAFT2jn