Q. 14 - An Introduction to General, Organic, and Biological Chemistry

Question

Automobile exhaust is a major cause of air pollution. One pollutant is the gas nitrogen oxide, which forms from nitrogen and oxygen caves in the air of the high temperatures in an automobile engine. Once emitted into the air, nitrogen oxide reacts with oxygen to produce nitrogen dioxide, a reddish brown gas with a sharp, pungent odor that makes up smog. One component of gasoline is heptane. ${\mathrm{C}}_7{\mathrm{H}}_{16+}$ a liquid which a density of$0.684$ g/mL. In one year, a typical automobile uses $550\mathrm{gal}$ of gasoline and produces 411 of nitrogen oxide. (7.4. 7.8.8.6)

a. Write balanced chemical equations for the production of nitrogen oxide and nitrogen dioxide.

b. If all the nitrogen oxide emitted by one automobile is converted to nitrogen dioxide in the atmosphere, how many kilograms of nitrogen dioxide are produced in one year by a single automobile?

c. Write a balanced chemical equation for the combustion of heptane.

d. Horn many moles of gaseous ${\mathrm{CO}}_2$are produced from the gasoline used by the typical automobile in one year, assuming the gasoline is all heptane?

e. How many liters of carbon dioxide at $TP$ are produced in one year from the gasoline used by the typical automobile?

Step-by-Step Solution

Answer

Part a)Nitrogen oxide combines with oxygen in the air and forms nitrogen dioxide.

$\begin{array}{l}2\mathrm{NO}\left(g\right)+{\mathrm{O}}_2\left(g\right)⟶2{\mathrm{NO}}_2\left(g\right)\\ \begin{array}{l}\text{ Nitrogen Oxygen Nitrogen }\\ \text{ oxide dioxide }\end{array}\end{array}$

Part b)620 moles of NO

Therefore, 620 moles of NO will produce 620 moles of ${\mathrm{NO}}_2$

Part c) The combustion reaction of heptane (a component of gasoline) is

${\mathrm{C}}_7{\mathrm{H}}_{16}\left(l\right)+11{\mathrm{O}}_2\left(g\right)⟶7{\mathrm{CO}}_2\left(g\right)+8{\mathrm{H}}_2\mathrm{O}\left(l\right)+\text{ energy }$

Heptane     Oxygen      Carbon dioxide   Water

Part d) Therefore, $9.9\times {10}^4$ moles of ${\mathrm{CO}}_2$ are produced from gasoline used by an automobile per year.

Part e) Therefore, $2.2\times {10}^6 \mathrm{L}$ of${\mathrm{CO}}_2$are produced from the gasoline used by an automobile in one year.

Step 1: Given information (Part a)

That are given equations are : (Part a)

Nitrogen oxide. NO, is produced from nitrogen and oxygen gases

Step 2: Explanation (Part a)

Nitrogen oxide, NO, is created from nitrogen and oxygen gasses within the discuss at tall temperatures. It is one of the poisons shaped in an vehicle motor at tall temperatures. The adjusted response for the generation of $\mathrm{NO}$ is

${\mathrm{N}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)\stackrel{\Delta }{⟶}2\mathrm{NO}\left(g\right)$

Nitrogen Oxygen Nitrogen oxide

Nitrogen oxide combines with oxygen in the air and forms nitrogen dioxide.

$2\mathrm{NO}\left(g\right)+{\mathrm{O}}_2\left(g\right)⟶2{\mathrm{NO}}_2\left(g\right)$

Nitrogen   Oxygen     Nitrogen

oxide                            dioxide

Step 3: Given information (Part b)

That are given equations are : (Part b)

The moles of NO can be calculated using the molar mass conversion factors.

Step 4: Explanation (Part b)

A common place car produces of nitrogen oxide. Let us change over the sum of nitrogen oxide radiated from Ib to grams. The balance and the change components can be composed as follows: 

$1\mathrm{lb}=454 \mathrm{g}$

$\frac{1\mathrm{lb}}{454 \mathrm{g}}$ and $\frac{454 \mathrm{g}}{1\mathrm{lb}}$

$41+\mathrm{bNO}\times \frac{454 \mathrm{gNO}}{{\mathrm{ltbNO}}_{\mathrm{NO}}}=18,614 \mathrm{gNO}$

Step 5: Given information (Part c)

That are given equations are: (Part c)

Let us convert moles of ${\mathrm{NO}}_2$ to grams using the molar mass conversion factors.

Step 6: Explanation (Part c)

Thus, a total of 620 moles of NO are emitted by one automobile in the atmosphere.

From the balanced reaction between nitrogen oxide and oxygen, we can see that 2 moles of NO produce 2 moles of ${\mathrm{NO}}_2$.

2 moles of $\mathrm{NO}=2$ moles of ${\mathrm{NO}}_2$

$\frac{2\text{ moles NO }}{2\text{ moles }{\mathrm{NO}}_2}$ and $\frac{2\text{ moles }{\mathrm{NO}}_2}{2\text{ moles NO }}$

620 moles $\mathrm{NO}\times \frac{2\text{ moles }{\mathrm{NO}}_2}{2\text{ moles }\mathrm{NO}}=620$ moles ${\mathrm{NO}}_2$

Therefore, 620 moles of NO will produce 620 moles of ${\mathrm{NO}}_2$.

Step 7: Explanation

Let us convert grams to kilograms of ${\mathrm{NO}}_2$

$1 \mathrm{kg}=1,000 \mathrm{g}$

Therefore, the conversion factor is

$$\begin{gathered} \frac{1 \mathrm{kg}}{1,000 \mathrm{g}}\text{ and }\frac{1,000 \mathrm{g}}{1 \mathrm{kg}} \\ 2.85\times {10}^4 \mathrm{g}\times \frac{1 \mathrm{kg}}{1,000 \mathrm{g}}=28.5 \mathrm{kg} \end{gathered}$$

Therefore, $28.5 \mathrm{kg}$ of ${\mathrm{NO}}_2$ is produced in one year by a single automobile.

Step 8: Explanation

(Part c) When a fuel is burnt in air, water and carbon dioxide gas are produced and energy is released in the form of heat or flame. The reaction is called a combustion reaction. The fuel is generally a hydrocarbon compound; it reacts with the oxygen in the air.

The combustion reaction of heptane (a component of gasoline) is

$$\begin{gathered} C_7{\mathrm{H}}_{16}\left(l\right)+11{\mathrm{O}}_2\left(g\right)⟶7{\mathrm{CO}}_2\left(g\right)+8{\mathrm{H}}_2\mathrm{O}\left(l\right)+\text{ energy } \\ \text{ Heptane Oxygen } \text{ Carbon dioxide Water } \end{gathered}$$

Step 9: Given information (Part d)

That are given equations are :(d).

Let us convert the volume of gasoline

Step 10: Explanation (Part d)

Let us convert the volume of gasoline used by an automobile per year from gallons to milliliters.

$$\begin{gathered} 1\mathrm{gal}=4\mathrm{qt} \\ 1\mathrm{qt}=946 \mathrm{mL} \\ 1\mathrm{gal}=4\mathrm{qt}\times \frac{946 \mathrm{mL}}{1\mathrm{qt}} \\ =3,784 \mathrm{mL} \end{gathered}$$

Thus, the conversion factors can be written as follows:

$$\begin{gathered} \frac{1\mathrm{gal}}{3784 \mathrm{mL}}\text{ and }\frac{3784 \mathrm{mL}}{1\mathrm{gal}} \\ 550\mathrm{gat}\times \frac{3784 \mathrm{mL}}{1\text{ gat }}=2.1\times {10}^6 \mathrm{mL}\text{ gasoline } \end{gathered}$$

Thus, the volume of gasoline used by an automobile per year is$2.1\times {10}^6 \mathrm{mL}$. Assuming the gasoline is all heptane, the volume of heptane used by an automobile is$2.1\times {10}^6 \mathrm{mL}$.

Step 11:Given information

That are given equations are : 

Let us convert the mass of heptane to moles of heptane $\left({\mathrm{C}}_7{\mathrm{H}}_{16}\right)$

Step 12: Explanation

We have the following:

$\text{ Density }=\frac{\text{ Mass }}{\text{ Volume }}$

By substituting the values of volume and density of heptane in the above expression, we get

Mass in $\mathrm{g}=0.684 \mathrm{g}/\mathrm{mL}\times 2.1\times {10}^6 \mathrm{mL}$

$=1.4\times {10}^6 \mathrm{g}$

Therefore, the mass of heptane is $1.4\times {10}^6 \mathrm{g}$.

Let us convert the mass of heptane to moles of heptane $\left({\mathrm{C}}_7{\mathrm{H}}_{16}\right)$using the molar mass conversion factors.

I mole of ${\mathrm{C}}_7{\mathrm{H}}_{16}=100 \mathrm{g}$ of ${\mathrm{C}}_7{\mathrm{H}}_{16}$

and$\frac{100 {\mathrm{gC}}_7{\mathrm{H}}_{16}}{1\text{ mole }{\mathrm{C}}_7{\mathrm{H}}_{16}}$

$1.4\times {10}^6{ \mathrm{g}}_{{\mathrm{gC}}_7{\mathrm{H}}_{16}}^{100+\frac{1\text{ mole }{\mathrm{C}}_7{\mathrm{H}}_{16}}{{ \mathrm{g}}_7{\mathrm{H}}_{16}}}=1.4\times {10}^4$ moles ${\mathrm{C}}_7{\mathrm{H}}_{16}$

Step 13: Given information

That are given sentences are:

From the combustion reaction, it can be stated that

Step 14 : Explanation

From the combustion reaction, it can be stated that

1 mole of ${\mathrm{C}}_7{\mathrm{H}}_{16}$ produces 7 moles of ${\mathrm{CO}}_2$.

I mole of ${\mathrm{C}}_7{\mathrm{H}}_{16}=7$ moles of ${\mathrm{CO}}_2$

$\frac{1\text{ mole }{\mathrm{C}}_7{\mathrm{H}}_{16}}{7\text{ moles }{\mathrm{CO}}_2}$ and $\frac{7\text{ moles }{\mathrm{CO}}_2}{1\text{ mole }{\mathrm{C}}_7{\mathrm{H}}_{16}}$

$1.4\times {10}^4$ moles ${\mathrm{C}}_7{\mathrm{H}}_{16}\times \frac{7\text{ moles }{\mathrm{CO}}_2}{1\text{ mele }{\mathrm{C}}_7{\mathrm{H}}_{16}}=9.9\times {10}^4$ moles ${\mathrm{CO}}_2$

$1.4\times {10}^4$ moles of ${\mathrm{C}}_7{\mathrm{H}}_{16}$ will produce $9.9\times {10}^4$ moles of ${\mathrm{CO}}_2$.

Therefore, $9.9\times {10}^4$moles of ${\mathrm{CO}}_2$ are produced from gasoline used by an automobile per year.

Step 15: Given information (Part e)

That are given equations are :(Part e)

Let us write the molar volume conversion factors.

Step 16: Explanation (Part e)

The moles of${\mathrm{CO}}_2$ produced from gasoline used by an automobile per year are $9.9\times {10}^4$. At STP (standard temperature and pressure), one mole of any gas occupies a volume of$22.4 \mathrm{L}$, which is called the molar volume.

Let us write the molar volume conversion factors.

1 mole of gas at STP = $22.4 \mathrm{L}$of gas

$\frac{1\text{ mole gas (STP) }}{22.4 \mathrm{L}\text{ gas }}$ and $\frac{22.4 \mathrm{L}\text{ gas }}{1\text{ mole gas(STP) }}$

Therefore, the volume occupled by$9.9\times {10}^4$ moles of ${\mathrm{CO}}_2$ gas can be calculated as

$9.9\times {10}^4$ moles ${\mathrm{CO}}_2\times \frac{22.4 {\mathrm{LCO}}_2}{1\text{ mele }{\mathrm{CO}}_2}=2.2\times {10}^6{ \mathrm{L}}_2$ of ${\mathrm{CO}}_2$gas

Therefore,$2.2\times {10}^6 \mathrm{L}$of ${\mathrm{CO}}_2$ are produced from the gasoline used by an automobile in one year.