Question

Butane is a hydrocarbon fuel obtained from crude oil. The equation for the complete combustion of butane is: 2 C4H10 + 13 O2 = 8 CO2 + 10 H2O2
14.5g of butane was burned in 72.0g of oxygen. Determine the limiting reagent. ​

Answer 1

Answer:

Solution

verified

Correct option is C)

Solution:- (C) C

4

H

10

(g)

+

2

13

O

2

(g)

⟶4CO

2

(g)

+5H

2

O

(l)

;Δ

c

H=−2658.0kJ/mol

The complete combustion of 1 mole of butane is represented by

C

4

H

10

(g)

+

2

13

O

2

(g)

⟶4CO

2

(g)

+5H

2

O

(l)

;Δ ccH=−2658.0kJ/mol

Δ

c

H should be negative and have a value of 2658kJ/mol.

Answer 2

Answer:

Solution

verified

Correct option is C)

Solution:- (C) C

4

H

10

(g)

+

2

13

O

2

(g)

⟶4CO

2

(g)

+5H

2

O

(l)

;Δ

c

H=−2658.0kJ/mol

The complete combustion of 1 mole of butane is represented by

C

4

H

10

(g)

+

2

13

O

2

(g)

⟶4CO

2

(g)

+5H

2

O

(l)

;Δ ccH=−2658.0kJ/mol

Δ

c

H should be negative and have a value of 2658kJ/mol.

Explanation: