Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasol

Question

ki77a

2 months ago

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Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasol

ine is 0.79 g/mL.

Chemistry

2 answers:

alisha [2.9K] · Answer 1 · 2026-03-03T19:21:15+03:00

Greetings!

Determine the mass (in kg) of water generated from burning 1.0 gallon (3.8 L) of gasoline (C8H18). Gasoline has a density of 0.79 g/mL.

First, we need to balance the reaction:

Information:

m (mass) =? (in Kg)

v (volume) = 3.8 L → 3800 mL

d (density) = 0.79 g/mL

MM (Molar Mass) of gasoline (C8H18)

C = 8*12 = 96 amu

H = 1*18 = 18 amu

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Thus, the Molar Mass (MM) of gasoline (C8H18) = 96 + 18 = 114 g/mol

Now, calculate the gasoline mass:

$d = \dfrac{m}{V}$

$0.79\:g/\diagup\!\!\!\!\!\!\!mL = \dfrac{m}{3800\:\diagup\!\!\!\!\!\!\!mL}$

$m_{gasoline} = 0.79\:g*3800$

$\boxed{m_{gasoline} = 3002\:g\:\:of\:\: C_{8} H_{18} }$

Next, find the mass of water:

Where: m (mass), MM (Molar Mass), n (mole count)

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$m_{water} = \dfrac{m_{gasoline}}{MM_{gasoline}} *n_{mol_{water}/1mol_{gasoline}}*MM_{H_{2}O}$

$m_{water} = \dfrac{3002}{114} *9*18$

$m_{water} = \dfrac{486324}{114}$

$m_{water} = 4266\:g\to\:m_{water} \approx 4.26\:Kg\to\:\boxed{\boxed{m_{water}\approx\:4.3\:Kg}}\end{array}}\qquad\quad\checkmark$

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I trust this is helpful, best regards... Dexteright02! =)

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lorasvet [2.7K] · Answer 2 · 2026-03-03T19:22:03+03:00

1) To start, we must establish the chemical equation for this combustion.

2C₈H₁₈ + 25O₂ ----> 16CO₂ + 18H₂O

2) Since we have the gasoline volume, we can convert it to grams using its density.

3.8 L is equal to 3800 mL.

To find grams: 3800 mL multiplied by 0.79 g/mL gives us 3002 grams.

3) We will convert grams into moles by utilizing the molar mass of gasoline.

The molar mass of C₈H₁₈ is calculated as (8 x 12.0) + (18 x 1.01) = 114.18 g/mol.

By converting, we find 3002 g is equal to 26.3 moles using (1 mol/ 114.18 grams).

4) We can now convert the moles of C₈H₁₈ to moles of water via the balanced equation.

26.3 moles of C₈H₁₈ is equal to 237 moles of H₂O (18 mol H₂O/ 2 moles C₈H₁₈).

5) We will convert moles to grams using the molar mass of water and then change this to kilograms.

The molar mass of H₂O is (2 x 1.01) + 16.0 = 18.02 g/mol.

Thus, 237 mol H₂O times (18.02 g/ 1 mol) gives us 4260 grams.

To convert to kilograms: 4260 grams equals 4.26 Kilograms.