Hello!
Calculate the mass (in kg) of water produced from the combustion of 1.0 gallon (3.8 L) of gasoline (C8H18). The density of gasoline is 0.79 g/mL.
Let's balance the equation:
Data:
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
--------------------------
MM (Molar Mass) of gasoline (C8H18) = 96 + 18 = 114 g/mol
Calculate the mass of gasoline:
[tex]d = \dfrac{m}{V}[/tex]
[tex]0.79\:g/\diagup\!\!\!\!\!\!\!mL = \dfrac{m}{3800\:\diagup\!\!\!\!\!\!\!mL}[/tex]
[tex]m_{gasoline} = 0.79\:g*3800[/tex]
[tex]\boxed{m_{gasoline} = 3002\:g\:\:of\:\: C_{8} H_{18} }[/tex]
Calculate the mass of water:
If: m (mass), MM (Molar Mass), n (number of mols)
Therefore:
[tex]m_{water} = \dfrac{m_{gasoline}}{MM_{gasoline}} *n_{mol_{water}/1mol_{gasoline}}*MM_{H_{2}O}[/tex]
[tex]m_{water} = \dfrac{3002}{114} *9*18[/tex]
[tex]m_{water} = \dfrac{486324}{114}[/tex]
[tex]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[/tex]
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I Hope this helps, greetings ... Dexteright02! =)
When gasoline is burnt, water and carbon dioxide are formed.
The equation of the combustion of gasoline is;
C8H18(g) + 25/2O2(g) -------->8CO2(g) + 9H2O(g)
We can obtain the mass of gasoline reacted as follows;
Density of gasoline = 0.79 g/mL
Mass of gasoline = ?
Volume of gasoline = 3.8 L or 3800mL
Density = mass/volume
mass = Density * volume
mass = 0.79 g/mL * 3800mL
mass = 3002 g
Amount of gasoline reacted = reacting mass/molar mass = 3002g/114g/mol
= 26.3 moles
From the reaction equation;
1 mole of gasoline yields 9 moles of water
26.3 moles gasoline yields 26.3 * 9/1 = 236.7 moles
Mass of water(in Kg) =236.7 moles * 18 g/mol/1000 = 4.26 Kg
Therefore, the amount of water produced = 4.26 Kg
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