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

Question

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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:

castortr0y [3K] · Answer 1 · 2026-03-24T21:56:00+03:00

Answer:

$m_{NO_2}=1424.16gNO_2$

Explanation:

Hello,

Initially, the combustion of octane is represented as indicated below:

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

Now, as 800 grams of octane combusts, we calculate the moles of oxygen that react using 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 reacting oxygen, we now examine the total oxygen used by:

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

Therefore, the 15% utilized in the NO₂ ultimately produces:

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

In the end, the grams of NO₂ generated are:

$O_2+2NO-->2NO_2$

Best regards.

lorasvet [2.7K] · Answer 2 · 2026-03-24T21:58:31+03:00

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 utilized for combusting octane

$87.54 mol O2 x \frac{15}{85}$ = 15.44 mol O2 used for generating NO2

O2 + 2NO → 2NO2

(15.44 mol O2) x (2/2) x (46.0056 g NO2/mol) = 710.33 g NO2