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IrinaK [193]
3 years ago
14

Hello :) for qn 3 , I got 0.01 mol but I’m not sure if it’s correct :/ I need help , thanks!

Chemistry
1 answer:
damaskus [11]3 years ago
8 0

Answer:

0.05 dm³

Explanation:

Please see the attached picture for full solution.

The question is asking for the volume of H₂SO₄, so we need to find the number of moles of KOH then the number of moles of H₂SO₄. (Using mole ratio from the balanced equation) Also, potassium hydroxide is KOH not K₂SO₄ :)

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Butane C4H10 (g),(Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2O(g) (Delta
shusha [124]

Answer: The enthalpy of combustion, per mole, of butane is -2657.4 kJ

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Putting the values we get :

\Delta H=[8\times H_f_{CO_2}+10\times H_f_{H_2O}]-[2\times H_f_{C_4H_{10}+13\times H_f_{O_2}}]

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