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Andreas93 [3]
3 years ago
6

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

.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.
Chemistry
1 answer:
shusha [124]3 years ago
3 0

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

Explanation:

The balanced chemical reaction is,

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

The expression for enthalpy change is,

\Delta H=[n\times H_f_{products}]-[n\times H_f_{reactants}]

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}}]

\Delta H=[(8\times -393.5)+(10\times -241.82)]-[(2\times -125.7)+(13\times 0)]

\Delta H=-5314.8kJ

2 moles of butane releases heat = 5314.8 kJ

1 mole of butane release heat = \frac{5314.8}{2}\times 1=2657.4kJ

Thus enthalpy of combustion per mole of butane is -2657.4 kJ

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                   Option-4 (3:2) is the correct answer.

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Answer:

47%

Explanation:

MgCO₃ (s) → MgO (s) + CO₂ (g)

CaCO₃ (s) → CaO (s) + CO₂ (g)

_______________________

MgCO₃ (s) + CaCO₃ (s) → MgO (s) + CaO (s) + 2CO₂ (g)

MgO: 40.3044 g/mol

MgCO₃: 84.3139 g/mol

CaO:56.0774 g/mol

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CO₂: 44.01 g/mol

15.42 g is the total mass of dolomite.

7.85 g is the sum of MgO and CaO produced.

This means that 7.57 g of CO₂ were produced.

44.01 g CO₂_____ 1 mol

7.57 g CO₂ _____  x

x = 0.172 mol CO₂

Considering the global reaction, we had 0.172/2 = 0.086 mol of MgCO₃ in the original sample.

1 mol MgCO₃ _______ 84.3139 g

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15.42 g dolomite ______ 100%

7.25 g MgCO₃ ________ z

z = 47%

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