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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

Verified Answer :

taskmasters

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 : 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.

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