'What mass of ethane gas (C2H6) must burn in excess oxygen to produce 2.00 x 103 kJ of heat, given the thermochemical equation'?
2016-03-30 9:23 am
Can you explain how to do this question for me? I'm really stuck the answer is 38.5 grams. I'm just not quite sure how they got there. Thanks
回答 (2)
2016-04-06 11:29 am
✔ 最佳答案
2C2H6(g) 7O2(g) → 4CO2(g) 6H2O(l) ΔH = -3.12 x 103 kJ 2 moles of C2H6 produces -3.12 x 10^3 kJ of heat
we have 2.00 x 10 ^ 3 kJ of heat so we need
2 * 2.00 x 10 ^ 3 / 3.12 x 10^3 moles of gas
= 1.28 moles or 1.28 moles * molar mass of the gas
= 1.28 mol * 30.0690 g/mol
= 38.5 g of ethane
2016-03-30 9:58 am
It is known that :
C₂H₆(g) + (5/2)O₂(g) → 2CO₂(g) + 3H₂O(l) .... ΔH = 1.56 × 10³ kJ
The combustion of 1 mole of C₂H₆ gives 1.56 × 10³ kJ of heat.
When 2.00 × 10³ kJ of heat is produced :
Number of moles of C₂H₆ burns = (2.00 × 10³ kJ) / (1.56 × 10³ kJ/mol) = 0.128 mol
Molar mass of C₂H₆ = 12.01×2 + 1.1×6 = 30.08 g/mol
Mass of C₂H₆ burns = (30.08 g/mol) × (0.128 mol) = 38.5 g
C₂H₆(g) + (5/2)O₂(g) → 2CO₂(g) + 3H₂O(l) .... ΔH = 1.56 × 10³ kJ
The combustion of 1 mole of C₂H₆ gives 1.56 × 10³ kJ of heat.
When 2.00 × 10³ kJ of heat is produced :
Number of moles of C₂H₆ burns = (2.00 × 10³ kJ) / (1.56 × 10³ kJ/mol) = 0.128 mol
Molar mass of C₂H₆ = 12.01×2 + 1.1×6 = 30.08 g/mol
Mass of C₂H₆ burns = (30.08 g/mol) × (0.128 mol) = 38.5 g
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