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

Answer:

187ppm.

Explanation:

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

$Mol_{gas} = \frac{m_{gas}}{M_{W}}$

$Mol_{gas} = \frac{16000g}{30}$

$Mol_{gas} = 533.3mol$

For the VOC's same:

$m_{voc} = 0.01*1000=10g$

then,

$Mol_{voc} = \frac{m_{voc}}{M_{W}}$

$Mol_{voc} = \frac{10}{100}$

$Mol_{voc} = 0.1mol$

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

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

$\xi = \frac{0.1}{0.1+533.3}*10^6$

$\xi = 187ppm.$