The source of oxygen that drives the internal combustion engine in an automobile

Question

The source of oxygen that drives the internal combustion engine in an automobile is air. Air is a mixture of gases, which are principally N2 (~79%) and O2 (~20%). In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas.

Part B

Both nitric oxide and nitrogen dioxide are pollutants that can lead to acid rain and global warming; collectively, they are called "NOx" gases. In 2004, the United States emitted an estimated 19 million short tons of nitrogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? Note that one short ton equals 2,000 lb. ?

Part C

The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO2 and water. If 85% of the oxygen in an engine is used to combust octane, and the remainder used to produce nitrogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

Explanation / Answer

given

one short ton equals 2,000 lb

1lb = 453.59gms

so

2000 lb = 907184.74 gms.

Now

19 million = 19,000,000

19,000,000 short tons = 19,000,000*2000lb

19,000,000*2000lb = 1.723*10^13 gms.

so NO2 is 1.723*10^13 gms.

Part C:

2 C8H18 + 25 O2....> 16 CO2 + 18 H2O

no.of moles of octane = 500 /114.23 = 4.4 moles

from the above balanced reation we can write
2 moles of octane is required 25 moles of O2
4.4 moles of octane is required ...........?
   = 25*4.4/2 = 55 mol O2 used to combust the octane
now we have to calculate no. of moles of O2 to combust 55moles of NO2

55 x (15/85) = 9.706 mol O2 used to form NO2

N2 + 2 O2 ....> 2 NO2

from the above balanced reation we can write
2 moles of O2 can give 2 moles of NO2
then 9.706 moles of O2 can give .....?
(9.706 mol O2) x (2/2) = 9.706 moles of NO2

weight of NO2 = mol.wt. of NO2*no.of moles of NO2
              = 9.706 * 46.005
              = 446.52 gms. of NO2

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