Question

In 1986 an electrical power plant in Taylorsville, Georgia, burned 8,376,726 \rm tons of coal, a national record at that time.

Assuming that the coal was 86.0 \% carbon by mass and that combustion was complete, calculate the number of tons of carbon dioxide produced by the plant during the year.


Assuming that the coal was 4.50 \% sulfur by mass and that combustion was complete, calculate the number of tons of sulfur dioxide produced by the plant during the year.

Answer 1

a)
`m_{CO_(2)} = 26,395,816.14\, tons`, b)
`m_{SO_(2)} = 753,105.939\,tons`

Step-by-step explanation:

a) Number of tons of carbon dioxide produced by the plant throughout the year:

A complete combustion means that a mole of
`CO_(2)` is produced by a mole of
`C` contained in coal. The yearly burnt carbon is:

`m_(C) = 0.86 \cdot 8,376,726\, tons\\m_(C) = 7,203,984.36\, tons`

The amount of the yearly
`CO_(2)` emission is obtained:

`m_{CO_(2)} = \frac{M_{CO_(2)}}{M_(C)}\cdot m_(C)`

`m_{CO_(2)} = (44.009\,(kg)/(kmol) )/(12.011 \,(kg)/(kmol) ) * 7,203,984.36\,tons`

`m_{CO_(2)} = 26,395,816.14\, tons`

b) Number of tons of tons of sulfur dioxide produced by the plant throughout the year:

The required information is found by applying the same approach seen on previous question. A complete combustion means that a mole of
`SO_(2)` is produced by a mole of
`S` contained in coal.

`m_(S) = 0.045 \cdot 8,376,726\, tons\\m_(S) = 376,952.67\, tons`

The amount of the yearly
`SO_(2)` emission is obtained:

`m_{SO_(2)} = \frac{M_{SO_(2)}}{M_(S)}\cdot m_(S)`

`m_{SO_(2)} = (64.062\,(kg)/(kmol) )/(32.065 \,(kg)/(kmol) ) * 376,952.67\, tons`

`m_{SO_(2)} = 753,105.939\,tons`