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Write the balanced equation of combustion and the side reaction where NO reacts with O2.
Relate both equations using the moles of O2. Balanced reactions will appear as:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
O2 + 2NO → 2NO2
Find the moles of O2 used upon combustion of 600g octane using stoichiometry. We need the molar mass of octane for this step
MM of C8H18
C - 12.01(8) = 96.08
H - 1.01(18) = 18.18
MM = 96.08 + 18.18 = 114.26 g/mol
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.
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 600 grams of octane.
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Our tutors have indicated that to solve this problem you will need to apply the Stoichiometry concept. You can view video lessons to learn Stoichiometry. Or if you need more Stoichiometry practice, you can also practice Stoichiometry practice problems.
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Based on our data, we think this problem is relevant for Professor West's class at UGA.