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HACTEHA
12 days 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:
castortr0y [2.8K]12 days ago
4 0

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 [2.5K]12 days 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 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

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