Question

Write the balanced equation for the combustion of isooctane (C₈H₁₈) to produce carbon dioxide and water. Use the smallest possible integers to balance the equation. Also, separate the + sign with 1 space.

2C₈H₁₈ + 25O₂ ? 16CO₂ + 18H₂O

You are correct. Your receipt no. is 161-2857

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Assuming gasoline is 90.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO₂ produced by the combustion of 1.42 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

g

Compare the number of moles of C8H18 to the number of moles of CO2 from the balanced reaction. Using this, the information given above, and the periodic table, set up an equation so the units cancel. HINT: The balanced equation has 25O2

Answer 1

Concepts and reason

Combustion reaction is a reaction in which, a hydrocarbon reacts with oxygen to produce carbon dioxide and water.

Fundamentals

Mass, m of a substance is related to number of moles, n as follows:

$m = n \times M$

Here, M is molar mass of substance.

The balanced chemical equation of combustion of isooctane $\left( {{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}\;} \right)$ is

$2{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}\; + {\rm{ }}25{{\rm{O}}_2}\;\longrightarrow{{}}{\rm{ }}16{\rm{C}}{{\rm{O}}_2}\; + {\rm{ }}18{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Volume of gasoline present (v)

$\begin{array}{c}\\v = \left( {1.42 \times {{10}^{10}}{\rm{gallon}}} \right)\left( {\frac{{3.785{\rm{ L}}}}{{1{\rm{ gallon}}}}} \right)\left( {\frac{{1000{\rm{ mL}}}}{{1{\rm{ L}}}}} \right)\\\\ = 5.37 \times {10^{13}}{\rm{ mL}}\\\end{array}$

Mass of gasoline (m)

$m = dv$

Where d is the density and v is the volume of gasoline

Substitute ${\rm{0}}{\rm{.69 g m}}{{\rm{L}}^{ - 1}}$ for d and $5.37 \times {10^{13}}{\rm{ mL}}$ for v

$\begin{array}{c}\\m = \left( {{\rm{0}}{\rm{.69 g m}}{{\rm{L}}^{ - 1}}} \right)\left( {5.37 \times {{10}^{13}}{\rm{ mL}}} \right)\\\\ = 3.7 \times {10^{13}}{\rm{ g}}\\\end{array}$

Now mass of isooctane $\left( {{m_i}} \right)$ will be

${m_i} = \frac{{90}}{{100}}\left( m \right)$

Substitute $3.7 \times {10^{13}}{\rm{ g}}$ for m

$\begin{array}{c}\\{m_i} = \frac{{90}}{{100}}\left( {3.7 \times {{10}^{13}}{\rm{ g}}} \right)\\\\ = 3.34 \times {10^{13}}{\rm{ g}}\\\end{array}$

Number of moles of isooctane

${n_i} = \frac{{{m_i}}}{M}$

Substitute $3.34 \times {10^{13}}{\rm{ g}}$ for ${m_i}$ and 114 g for M

$\begin{array}{c}\\{n_i} = \frac{{3.34 \times {{10}^{13}}{\rm{ g}}}}{{114{\rm{ g}}}}\\\\ = 0.029 \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ mol}}\\\end{array}$

According to equation, 2 moles isooctane gives 18 moles carbon dioxide

Hence 0.029 mol isooctane will give ${n_{{\rm{C}}{{\rm{O}}_2}}}$ mol

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_2}}} = \frac{{18}}{2}\left( {0.029 \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ mol}}} \right)\\\\ = 0.26 \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ mol}}\\\end{array}$

Hence theoretical yield of carbon dioxide in moles is $0.26 \times {\rm{1}}{{\rm{0}}^{13}}$ mol

Theoretical yield of carbon dioxide in g $\left( {{m_{{\rm{C}}{{\rm{O}}_2}}}} \right)$

${m_{{\rm{C}}{{\rm{O}}_2}}} = {n_{{\rm{C}}{{\rm{O}}_2}}} \times {M_{{\rm{C}}{{\rm{O}}_2}}}$

Substitute $0.26 \times {\rm{1}}{{\rm{0}}^{13}}$ mol for ${n_{{\rm{C}}{{\rm{O}}_2}}}$ and 44 g for ${M_{{\rm{C}}{{\rm{O}}_2}}}$

$\begin{array}{c}\\{m_{{\rm{C}}{{\rm{O}}_2}}} = 0.26 \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ mol}} \times 44{\rm{ g}}\\\\{\rm{ = 11}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ g}}\\\end{array}$

According to equation, 2 moles isooctane gives 18 moles carbon dioxide

Hence ${n_i}$ isooctane will give ${n_{{\rm{C}}{{\rm{O}}_2}}}$ mol

$\begin{array}{c}\\{n_{{\rm{C}}{{\rm{O}}_2}}} = \frac{{18}}{2}\left( {{n_i}} \right)\\\\ = 9{n_i}\\\end{array}$

Ans:

Hence, the balanced equation is: $2{{\rm{C}}_{\rm{8}}}{{\rm{H}}_{{\rm{18}}}}\; + {\rm{ }}25{{\rm{O}}_2}\;\longrightarrow{{}}{\rm{ }}16{\rm{C}}{{\rm{O}}_2}\; + {\rm{ }}18{{\rm{H}}_{\rm{2}}}{\rm{O}}$

Theoretical yield of carbon dioxide is ${\rm{11}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{13}}{\rm{ g}}$