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r-ruslan [8.4K]
2 years ago
14

Octane has a density of 0.692 g/ml at 20oc. how many grams of o2 are required to burn 15.0 gal of c8h18 (the average capacity of

a fuel tank. 1 gallon = 3.78541 l)?
Chemistry
1 answer:
Hatshy [7]2 years ago
4 0

Octane has a density of 0.692 g/ml at 20°C. Grams of O₂ required to burn 15.0 gal of C₈H₁₈ is 1.37868×10⁵g.

Octane is a hydrocarbon having eight carbon atoms and have single bonds only.

Given,

Density of octane = 0.692g/ml

Temperature = 20°C = 293K

Amount of Octane present = 15gal

Molar mass of octane = 114g

We know that,

1 gal = 3.78541L = 3785.4ml

Hence, 15 gal = 56781 ml

Now let's calculate the Mass of octane required:

Mass of octane = 0.692 x 56781

Mass of octane = 39292.45 g

According to the given equation,

C₈H₁₈ + 12.5O₂ ---------> 8CO₂ + 9H₂O

Also, 114g of octane needs-------> 400g of oxygen

39292.45g of octane needs ---------> 137868.24g of oxygen

Hence, oxygen needed to burn octane is 1.37868×10⁵g

Learn more about Octane here, brainly.com/question/21268869

#SPJ4

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aleksandr82 [10.1K]

Answer:

5 litres of solution are requried

Explanation:

Given that the typical units of

concentration

are

m

o

l

⋅

L

−

1

, to get

concentration

or

amount of moles

or

volume of solution

we just have to take the appropriate product or quotient.

We want

1.0

⋅

m

o

l

N

a

O

H

:

this the product

Volume of solution

×

Concentration

.

So we need the quotient:

Moles of solute

Concentration

=

1.0

⋅

m

o

l

0.2

⋅

m

o

l

⋅

L

−

1

=

5

⋅

L

.

(i.e.

1

1

⋅

L

−

1

=

1

⋅

L

)

Note that we go to such trouble in including the units in these calculations as an extra check on our arithmetic. Sometimes you ask yourself should I divide or should I multiply. Dimensional analysis answers our question. We wanted an answer in

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