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
1 answer:
156251.099 grams of O₂ are required to burn 17.0 gal of C₈H₁₈
<h3>Further explanation</h3>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:
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 O₂ = 4882.846
grams O₂ = mole x molar mass
grams O₂ = 4882.846 x 32
grams O₂ = 156251.099
<h3>Learn more </h3>moles of water
grams of oxygen required
mass of copper required
Keywords: Octane, mole, mass, gal, density
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