Question

How many liters of carbon dioxide are released when 355 grams of pentane (`C_5H_10`) is burned in air at STP?

Answer 1

`566.9" L of "CO_2`

Explanation:

The balanced equation is:

`2C_5H_10 + 15O_2 rarr 10CO_2 + 10H_2O`

We begin Dimensional Analysis by writing the given as a fraction over 1:

`(355" g of "C_5H_10)/1`

Multiply by the conversion factor from g of `C_5H_10` to mol of `C_5H_10`:

`(355" g of "C_5H_10)/1(1" mol of "C_5H_10)/(70.13" g of "C_5H_10)`

Multiply by the conversion factor from moles of `C_5H_10` to moles of `CO_2`, obtained from the balanced equation:

`(355cancel(" g of "C_5H_10))/1(1" mol of "C_5H_10)/(70.13cancel(" g of "C_5H_10))(10" mol of "CO_2)/(2" mol of "C_5H_10)`

Multiply by the gas volume at STP conversion factor:

`(355cancel(" g of "C_5H_10))/1(1cancel(" mol of "C_5H_10))/(70.13cancel(" g of "C_5H_10))(10" mol of "CO_2)/(2cancel(" mol of "C_5H_10))(22.4" L of "CO_2)/(1" mol of "CO_2)`

Remove all of the cancelled units and perform the calculation:

`355/70.13(10)/(2)(22.4" L of "CO_2) = 566.9" L of "CO_2`

Answer 2

`60l`
Calculate moles of hydrocarbon, thus moles of gas and volume of gas produced.

Explanation:

First, we can work out the number of moles of hydrocarbon.

Note
You wrote pentANE, which is `C_5H_12`, but then gave the formula for pentENE, which is `C_5H_10`. I'm going to assume you meant pentene for the sake of this example, and, while the numbers will be different, the method will be the same.

We can work out the number of moles of pentene by the formula

`Mass=M_r xx mol`
(if you think of Mr Mole, you'll never forget this formula)
`mol=(Mass)/M_r`
`mol=355/(5(12)+10)`
`mol=355/70`
`mol=5.071...`
So we have 5 moles of pentene.

Now, we want to know how much carbon dioxide is produced.

The unbalanced equation is:

`C_5H_10 + O_2 -> CO_2 + H_2O`

From the conservation of mass, the number of carbon atoms on each side must be equal. This means both the LHS and RHS must have 5 carbons. The RHS only has 1 at the moment, so we multiply this by 5 to get:

`C_5H_10 + O_2 -> 5CO_2 + H_2O`
This still isn't balanced, but we don't need to balance it.

This gives us a ratio;
`C_5H_10:CO_2=1:5`
And since we have 5 moles of pentene:
`C_5H_10:CO_2=5:25`
And we have 25 moles of `CO_2`.

Now, the most important part of all:
One mole of any ideal gas at standard temperature and pressure takes up a space of `24dm^3`

`1mol:24dm^3`
But we have 5 moles of `CO_2`
`5mol:60dm^3`
And we have `60dm^3` of `CO_2`

1 decimeter cubed is equal to 1 litre. So, we have 60 litres of carbon dioxide produced.