Question

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.

Answer 1

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

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

Calculate the mass of gasoline:

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

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

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

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

Calculate the mass of water:  

If: m (mass), MM (Molar Mass), n (number of mols)  

Therefore:

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

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

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

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

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I Hope this helps, greetings ... Dexteright02! =)

Answer 2

1) first, we need to write down the reaction equation for this combustion. 

2C₈H₁₈ + 25O₂ ----> 16CO₂ + 18H₂O

2) we have the volume of gasoline, we can use the density to convert to grams

3.8 L = 3800 mL

3800 mL (0.79 g/ 1 mL)= 3002 grams

3) let's convert the grams to moles using the molar mass of gasoline

molar mass C₈H₁₈= (8 x 12.0) + (18 x 1.01)= 114.18 g/mol

3002 g (1 mol/ 114.18 grams) = 26.3 moles

4) let's use the balanced equation to convert moles of C₈H₁₈ to moles of water.

26.3 moles C₈H₁₈ (18 mol H₂O/ 2 moles C₈H₁₈)= 237 moles H₂O

5) finally, let's convert the moles to grams using the molar mass of water and then to Kg

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

237 mol H₂O (18.02 g/ 1 mol)= 4260 grams

4260 grams= 4.26 Kilograms