Question

F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

Answer 1

Answer : The mass of
`CO_2` produced will be, 429.264 grams

Explanation : Given,

Mass of
`C_4H_8` = 136.6 g

Molar mass of
`C_4H_8` = 56 g/mole

Molar mass of
`CO_2` = 44 g/mole

First we have to calculate the moles of
`C_4H_8`.

`\text{Moles of }C_4H_8=\frac{\text{Mass of }C_4H_8}{\text{Molar mass of }C_4H_8}=(136.6g)/(56g/mole)=2.439moles`

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

The balanced chemical reaction is,

`C_4H_8+6O_2\rightarrow 4CO_2+4H_2O`

From the balanced reaction we conclude that

As, 1 mole of
`C_4H_8` react to give 4 moles of
`CO_2`

So, 2.439 moles of
`C_4H_8` react to give
`(4)/(1)* 2.439=9.756` 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`

`\text{Mass of }CO_2=(9.756mole)* (44g/mole)=429.264g`

Therefore, the mass of
`CO_2` produced will be, 429.264 grams.

Answer 2

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol C4H8) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2