Question

Assume the gasoline is burned with 99% efficiency in a car engine, with 1% remaining unburned in the exhaust gases as VOCs (volatile organic compounds). If the engine exhausts 16 kg of gases (MW=30) for each kg of gasoline (MW=100), calculate the fraction of VOCs in the exhaust. Give your answer in parts per million

Answer 1

187ppm.

Step-by-step explanation:

To develop the problem we first need to perform mass conversions to moles. In this way for gases your conversion is given by,

`Mol_(gas) = (m_(gas))/(M_(W))`

`Mol_(gas) = (16000g)/(30)`

`Mol_(gas) = 533.3mol`

For the VOC's same:

`m_(voc) = 0.01*1000=10g`

then,

`Mol_(voc) = (m_(voc))/(M_(W))`

`Mol_(voc) = (10)/(100)`

`Mol_(voc) = 0.1mol`

We only need to obtain the fraction of Voc's in exhaust. This will be in particles per million, so

`\xi = \fra{Mol_(voc)}{(Mol_(gas)+Mol_(voc))}`

`\xi = (0.1)/(0.1+533.3)*10^6`

`\xi = 187ppm.`