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ASHA 777 [7]
3 years ago
13

Butane C4 H10 (g),(Delta.Hf = –125.7), combusts in the presence of oxygen to form CO2 (g) (Delta.Hf = –393.5 kJ/mol), and H2 O(g

) (Delta.Hf = –241.82) in the reaction:
2 upper C subscript 4 upper H subscript 10 (g) plus 13 upper O subscript 2 (g) right arrow 8 upper C upper O subscript 2 plus 10 upper H subscript 2 upper O (g).

What is the enthalpy of combustion, per mole, of butane?
Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants..
–5,314.8 kJ/mol
–2,657.4 kJ/mol
2,657.4 kJ/mol
5,314.8 kJ/mol
Chemistry
2 answers:
a_sh-v [17]3 years ago
8 0

Answer:

B: –2,657.4 kJ/mol

Explanation:

took the test

Reptile [31]3 years ago
4 0

Answer:

  • <u>Second choice:</u>

                       \Delta H_{rxn}=-2,657.4kJ/mol

Explanation:

<u>1. Combustion reaction </u>

      2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O

<u>2. Enthalpy of reaction equation</u>

     \Delta H_{rxn}=\sum\Delta H_f(products)-\sum\Delta H_f(reactants)

<u>3. Substitute and divide by the coefficient of butane that appears in the combustion reaction</u>

      \Delta H_{rxn}=\dfrac{8(-393.5kJ/mol)+10(-241.82kJ/mol)-2(-125.7kJ/mol)}{2mol}

      \Delta H_{rxn}=-2,657.4kJ/mol\leftarrow answer

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