Question

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

Answer 1

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:

$\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}$

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

$mole\:O_2\:=\:\frac{25}{2} \times\:390.627$

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