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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.

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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nataly862011 [7]

Answer:

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3 years ago

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-----------------------------

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4 0

3 years ago

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9. Thallium-208 has a half-life of 3.053 min. How long will it take for 120 g of it to decay

damaskus [11]

Answer:

12.213 minutes will be taken for 120 g-Thalium-208 to decay to 75 grams.

Explanation:

Radioactive isotopes decay exponentially in time, the mass of the isotope ( $m(t)$ ), in grams, is described by the formula in time ( $t$ ), in minutes:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ (1)

Where:

$m_{o}$ - Initial mass of the isotope, in grams.

$\tau$ - Time constant, in minutes.

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$\tau = \frac{t_{1/2}}{\ln 2}$ (2)

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$\tau = \frac{t_{1/2}}{\ln 2}$

$\tau = \frac{3.053\,min}{\ln 2}$

$\tau \approx 4.405\,min$

$t = -\tau \cdot \ln \frac{m(t)}{m_{o}}$

$t = -(4.405\,min)\cdot \ln \left(\frac{7.5\,g}{120\,g} \right)$

$t \approx 12.213\,min$

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3 years ago