Question

What mass of CO2 is produced from the combustion of 2.00 gallons of gas? Assume the gas is all octane (C8H18) with a density of 0.703 g/mL. 1 gallon = 3785 mL

Answer 1

Answer : The mass of
`CO_2` produced form the combustion is, 16.43 kg

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

`Density=(Mass)/(Volume)`

First we have to calculate the mass of octane.

Given :

Density of octane = 0.703 g/mL

Volume = 2.00 gallons = 7570 mL

conversion used : 1 gallon = 3785 mL

Now put all the given values in the above formula, we get:

`0.703g/mL=(Mass)/(7570mL)`

`Mass=5321.71g`

Now we have to calculate the moles of octane.

`\text{Moles of octane}=\frac{\text{Mass of octane}}{\text{Molar mass of octane}}`

Molar mass of octane = 114 g/mole

`\text{Moles of octane}=(5321.71g)/(114g/mole)=46.68mole`

Now we have to calculate the moles of
`CO_2`.

The balanced chemical combustion reaction of octane will be:

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

From the balanced chemical reaction we conclude that:

As, 2 moles of
`C_8H_(18)` react to give 16 moles of
`CO_2`

So, 46.68 moles of
`C_8H_(18)` react to give
`(46.68)/(2)* 16=373.44` moles of
`CO_2`

Now we have to calculate the mass of
`CO_2`.

`\text{ Mass of }CO_2=\text{ Moles of }CO_2* \text{ Molar mass of }CO_2`

Molar mass of
`CO_2` = 44 g/mol

`\text{ Mass of }CO_2=(373.44moles)* (44g/mole)=16431.36g=16.43kg`

Thus, the mass of
`CO_2` produced form the combustion is, 16.43 kg