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igor_vitrenko [27]
3 years ago
11

The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into CO

2 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 800 grams of octane.
Chemistry
2 answers:
borishaifa [10]3 years ago
4 0

Answer:

710,33 g NO2

Explanation:

2 C8H18 + 25 O2 → 16 CO2 + 18 H2O  

(800 g octane) / (114.2293 g C8H18/mol x (25/2)) = 87.54 mol O2 used to combust the octane

87.54 mol O2 x \frac{15}{85} = 15.44 mol O2 used to form NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710,33 g NO2

Brilliant_brown [7]3 years ago
4 0

Answer:

m_{NO_2}=1424.16gNO_2

Explanation:

Hello,

At first, the combustion of octane is illustrated as shown below:

C_8H_{18}(g)+\frac{25}{2}O_2(g)-->8CO_2(g)+9H_2O(g)

Now, since 800 grams of octane are burned, we compute the moles of oxygen that reacted via stoichiometry:

n_{O_2}=800gC_8H_{18}*\frac{1molC_8H_{18}}{114gC_8H_{18}}*\frac{\frac{25}{2}molO_2}{1molC_8H_{18}} \\n_{O_2}=87.72molO_2

As that is the combusting oxygen, we now look for the total oxygen that was employed by:

n_{O_2}^{tot}=\frac{87.72molO_2}{0.85} =103.2molO_2

Therefore the 15% used in the NO₂ turns out:

n_{O_2}^{forNO_2}=103.2molO_2*0.15=15.48molO_2

Finally, for the NO₂ production:

O_2+2NO-->2NO_2

The produced grams of NO₂ are:

m_{NO_2}=15.48molO_2*\frac{2molNO_2}{1molO_2}*\frac{46gNO_2}{1molNO_2}\\\\ m_{NO_2}=1424.16gNO_2

Best regards.

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