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A sample of liquid heptane (C7H16) weighing 11.50g is reacted with 1.300 moles of oxygen gas. The heptane is burned completely. After the reaction is complete, the amount of gas present is 1.050 moles.

a. How many moles of CO are produced?

b. How many moles of CO2 are produced?

c. How many moles of O2 are left over?

Respuesta :

Answer:

Explanation:

2C₇H₁₆ + 15 O₂ = 14 CO + 16 H₂O

C₇H₁₆ +  11O₂ = 7 CO₂ + 8 H₂O

Let the moles of CO formed be x .

moles of heptane taken

= 11. 5 / mol weight

= 11.5 / 100

= .115 moles

moles of C₇H₁₆  reacting to form CO = x / 7 moles

moles of C₇H₁₆ reacting to form CO₂ = (.115 - x / 7 )

moles of oxygen reacted to form CO = (15 / 14) x

moles of oxygen reacted to form CO₂ = 11 (.115 - x / 7 )

Total moles of oxygen reacted =  (15 / 14) x + 11 (.115 - x / 7 )

= 1.071 x +1.265 - 1.57 x

= 1.265 - .499 x

moles of oxygen  remaining unreacted

= 1.3 -  1.265 + .499 x

= .035 + .499 x  

Total moles of gases after reaction

= .035 + .499 x  + x + 7  (.115 - x / 7 ) = 1.05

.035 + .499 x +x + .805 - x = 1.05

.84 +.499 x  = 1.05

.499 x = .21

x = .42 moles

So moles of CO formed = .42 moles

b )

moles of CO₂ formed =  7  (.115 - x / 7 )

= .805 - .42

= .385 moles

c ) moles of O₂ left over =

.035 + .499 x  

= .035 + .499 x .42

= .244 moles

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