Octane has a density of 0.692 g/ml at 20∘c. how many grams of o2 are required to burn 17.0 gal of c8h18

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mrbear7147 mrbear7147

05-10-2017
Chemistry

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Octane has a density of 0.692 g/ml at 20∘c. how many grams of o2 are required to burn 17.0 gal of c8h18

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ardni313 ardni313 · Answer 1 · 2019-09-04T10:28:20+02:00

156251.099 grams of O₂ are required to burn 17.0 gal of C₈H₁₈

Further explanation

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g / cm³ or kg / m³

Density formula:

[tex]\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}[/tex]

ρ = density

m = mass

v = volume

1 gal equal to = 3785.41 ml

then 17.0 gal = 17 x 3785.41 = 64351.97 ml Octane

grams Octane = ρ x ml

grams Octane = 0.692 g.ml x 64351.97

grams Octane = 44531.563

molar mass Octane (C₈H₁₈) = 114

mole Octane = grams : molar mass

mole Octane = 44531.563 : 114

mole Octane = 390.627

From the reaction

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9H₂O

mole C₈H₁₈ : mole O₂ = 1 : 25/2

[tex]mole\:O_2\:=\:\frac{25}{2} \times\:390.627[/tex]

mole O₂ = 4882.846

grams O₂ = mole x molar mass

grams O₂ = 4882.846 x 32

grams O₂ = 156251.099

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Keywords: Octane, mole, mass, gal, density

Cricetus Cricetus · Answer 2 · 2021-09-29T13:02:30+02:00

"[tex]1.56\times 10^5 \ g[/tex]" of [tex]O_2[/tex] are required to burn 17.0 gal of [tex]C_8 H_{18}[/tex].

According to the question,

Density of Octane = 0.692 g/mL
Temperature = 20°C

We know that,

[tex]1 \ gal = 3785.4 \ mL[/tex]

then,

[tex]17 \ gal = 64352 \ mL[/tex]

Now,

→ Mass of Octane will be:

= [tex]0.692\times 64352[/tex]

= [tex]44531 \ g[/tex]

or,

→ [tex]114 \ g \rightarrow 400 \ g[/tex]

→ [tex]44531 \ g \rightarrow \frac{400\times 44531}{114}= 156251 \ g[/tex]

hence,

The Oxygen (O₂) needed will be:

= [tex]1.56\times 10^5 \ g[/tex]

Thus the above answer is appropriate.

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