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?

Answer 1

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

Answer 2

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 $C_4H_8$.

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

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 $\frac{4}{1}\times 2.439=9.756$ moles of $CO_2$

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

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

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

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