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How many kg of CO2 are produced when 1.00 galloon of octane (C8H18, density = 0.703 g/mL) is completely combusted in the presence of O2 to form CO2 and H2O?

Question

How many kg of CO2 are produced when 1.00 galloon of octane (C8H18, density = 0.703 g/mL) is completely combusted in the presence of O2 to form CO2 and H2O?

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Step 1

Density of a substance is the ratio of its mass and volume. It is expressed in equation (1) in which d is the density, m is the mass and V is the volume.

 

The relation between gallon (gal) and millilitres (mL) is given in equation (2).

The number of moles is the ratio of given mass by molar mass. It is expressed in equation (3) in which n is the number of moles, m is the given mass and M is the molar mass.

= P
V
(1)
1 gal 3785.41 mL
(2)
. (3)
М
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= P V (1) 1 gal 3785.41 mL (2) . (3) М

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Step 2

The balanced reaction of complete combustion of octane is written below:

2C8H18(g)+25O2(g) →16CO2(g)+18H2O(g)

From equation (2), 1 gal of octane is equal to 3785.41 mL . The mass of octane is calculated by substituting 3785.41 mL for V and 0.703 g/mL for d in equation (1).

0.703 g/m3785.41 mL
m 2661.1 g
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0.703 g/m3785.41 mL m 2661.1 g

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Step 3

The number of moles of octane is calculated is shown in equation (4) by usin...

2661.1 g
octane
114.23 g/mol
(4)
=23.3 mol
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2661.1 g octane 114.23 g/mol (4) =23.3 mol

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