Question

Assuming gasoline is 89.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO2 produced by the combustion of 1.80 x 1010 gallons of gasoline (the estimated annual consumption of gasoline in the U.S.)? Remember, there are 3.785 liters in 1 gallon and assume that isooctane is the only carbon containing component of gasoline. Scientific notation can be entered as follows: 1.23 x 1023 = 1.23E23

Answer 1

Answer: The theoretical yield of carbon dioxide is $1.453\times 10^{14}g$

Explanation:

We are given:

Volume of isooctane = $1.80\times 10^{10}gallons$

To convert this into liters, we use the conversion factor:

1 gallon = 3.785 L

So, $1.80\times 10^{10}gallon\times (\frac{3.785L}{1gallon})=6.813\times 10^{10}L$

To calculate the mass of isooctane, we use the equation:

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

Volume of isooctane = $6.813\times 10^{10}L=6.813\times 10^{13}mL$    (Conversion factor:  1 L = 1000 mL)

Density of isooctane = 0.692 g/mL

Putting values in above equation, we get:

$0.692g/mL=\frac{\text{Mass of isooctane}}{6.813\times 10^{13}mL}\\\\\text{Mass of isooctane}=(0.692g/mL\times 6.813\times 10^{13}mL)=4.714\times 10^{13}g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

Given mass of isooctane = $4.714\times 10^{13}g$

Molar mass of isooctane = 114.22 g/mol

Putting values in equation 1, we get:

$\text{Moles of isooctane}=\frac{4.714\times 10^{13}g}{114.22g/mol}=4.127\times 10^{11}mol$

The chemical equation for the combustion of isooctane follows:

$2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O$

By Stoichiometry of the reaction;

2 moles of isooctane produces 16 moles of carbon dioxide.

So, $4.127\times 10^{11}mol$ of isooctane will produce = $\frac{16}{2}\times 4.127\times 10^{11}mol=3.3016\times 10^{12}mol$ of carbon dioxide

• Now, calculating the mass of carbon dioxide from equation 1, we get:

Molar mass of carbon dioxide = 44.00 g/mol

Moles of carbon dioxide = $3.3016\times 10^{12}mol$

Putting values in equation 1, we get:

$3.3016\times 10^{12}mol=\frac{\text{Mass of carbon dioxide}}{44.00g/mol}\\\\\text{Mass of carbon dioxide}=(3.3016\times 10^{12}mol\times 44.00g/mol)=1.453\times 10^{14}g$

Hence, the theoretical yield of carbon dioxide is $1.453\times 10^{14}g$

Answer 2

Based on the data provided, the theoretical yield of CO2 is 4.27 10^13 kg

What is theoretical yield?

Theoretical yield of a reaction is the amount of products obtained from the equation of reaction.

The theoretical yield of CO2 is obtained from the equation of the reaction given below:

2 C8H18 + 25 O2 ----> 16 CO2 + 18 H2O

molar mass of C8H18 = 114 g

molar mass of CO2 = 44 g

molar volume at STP = 22.4 L

• 2 × 114g of C8H18 produces 16 × 44 g of CO2
• 228 g of C8H18 produces 224 g of CO2

Volume of gasoline in litres = 1.80 × 10^10 × 3.785

Volume of gasoline = 6.813 × 10^10 L

Density of gasoline = 0.692 g/mL = 692 kg/L

mass of gasoline = 6.813 × 10^10 L × 692 kg/L

mass of gasoline = 4.714 × 10^13 kg

mass of gasoline in 89% solution = 0.89 × 4.714 × 10^13 kg= 4.195 10^13 kg of gasoline.

Theoretical yield of CO2 = 4.195 × 10^13 kg × 0.224/0.228

Theoretical yield of CO2 = 4.27 10^13 kg of CO2

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