Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 19.70 mpg, you would use approximately 50.76 gallons of gasoline every month. If gasoline is approximated as C8H18 (density = 0.703 g/mL)...?

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Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 19.70 mpg, you would use approximately 50.76 gallons of gasoline every month. If gasoline is approximated as C8H18 (density = 0.703 g/mL)...?

How many grams of carbon dioxide does your vehicle emit every month?

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Answer 1 · 2017-10-17T02:19:55

Each month, the car will produce 417 120 g of $C O_{2}$ $CO_2$

Explanation:

First, let's get all the unit sorted out.

Assuming this is a US gallon, it will be equivalent to 3785 mL, and since the density of octane (representing the gasoline) is 0.703 g/mL, the one gallon would be

$3785 m L \times 0.703 \frac{g}{m L} = 2661 g$ $3785 mLxx0.703g/(mL)=2661 g$ per gallon

You have used 50.76 gallons each month, so

$50.76 \times 2661 = 135072$ $50.76xx2661 = 135 072$ grams each month.

The reaction that produces $C O_{2}$ $CO_2$ is

$2 C_{8} H_{18} + 25 O_{2} \to 16 C O_{2} + 18 H_{2} O$ $2C_8H_18+25O_2rarr16CO_2+18H_2O$

so, we need to change those grams of gasoline into moles of octane:

Moles = $135072 g \div 114 \frac{g}{mol} = 1185 mol$ $135 072g-:114g/"mol" = 1185 "mol"$

According to the reaction, we get 8 moles of $C O_{2}$ $CO_2$ for every 1 mole of $C_{8} H_{18}$ $C_8H_18$ , so the number of moles of $C O_{2}$ $CO_2$ produced will be

$1185 \times 8 = 9480$ $1185xx8=9480$ moles of $C O_{2}$ $CO_2$

Since each mole of $C O_{2}$ $CO_2$ has a mass of 44.0 g, the total monthly output of $C O_{2}$ $CO_2$ will be...

$9480 mol \times 44.0 \frac{g}{mol} = 417120$ $9480 "mol" xx 44.0 g/"mol"=417 120$ g or 417.1 kg

Answer 2 · 2017-10-17T02:56:05

About $4.171 \times 10^{5}$ $4.171xx10^5$ grams of $C O_{2}$ $CO_2$ is emitted.

Explanation:

[Step1] First, calculate the volume of gasoline $V$ $V$ to drive 1000 miles.

$V = 50.76$ $V= 50.76$ (gallons)
Assuming gallon to be the US gallon( $1$ $1$ US gal = $3.785$ $3.785$ L),
$V = 50.76 \cdot 3.785 = 192.1$ $V=50.76 * 3.785 = 192.1$ (L)

[Step2] Then, calculate the mass(g) and amount of substance(mol) of $C_{3} H_{8}$ $C_3H_8$ (octane).

The mass is $192.1 \cdot 10^{3}$ $192.1 * 10^3$ (mL) $\cdot 0.703$ $* 0.703$ (g/mL) $= 1.351 \times 10^{5}$ $= 1.351xx10^5$ (g).

Since the molar mass of $C_{8} H_{18}$ $C_8H_18$ (octane) is $12 \cdot 8 + 1 \cdot 18 = 114$ $12*8+1*18=114$ (g/mol) , the amount of substance of octane is $\frac{1.351 \times 10^{5}}{114} = 1.185 \times 10^{3}$ $(1.351xx10^5)/114= 1.185xx10^3$ (mol)

[Step3] Calculate the $C O_{2}$ $CO_2$ emission.
The chemical equation for burning $C_{8} H_{18}$ $C_8H_18$ is
$2 C_{8} H_{18} + 25 O_{2} \to 16 C O_{2} + 18 H_{2} O$ $2C_8H_18 + 25O_2 -> 16CO_2 + 18H_2O$ . Eight moles of carbon dioxcide is emitted per one mole of octane.

Thus the amount of substance for $C O_{2}$ $CO_2$ is $1.185 \times 10^{3} \cdot 8 = 9.480 \cdot 10^{3}$ $1.185xx10^3*8=9.480*10^3$ (mol).
The molar mass of $C O_{2}$ $CO_2$ is $12 \cdot 1 + 16 \cdot 2 = 44$ $12*1+16*2 =44$ .
Therefore, the mass of $C O_{2}$ $CO_2$ emitted in a month is $9.480 \times 10^{3} \cdot 44 = 4.171 \times 10^{5}$ $9.480xx10^3*44 =4.171xx10^5$ (g).

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Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 19.70 mpg, you would use approximately 50.76 gallons of gasoline every month. If gasoline is approximated as C8H18 (density = 0.703 g/mL)...?

How many grams of carbon dioxide does your vehicle emit every month?

2 Answers

Explanation:

Explanation:

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