Question

part a
the combustion of gasoline produces carbon dioxide and water. assume gasoline to be pure octane (c8h18) and calculate how many kilograms of carbon dioxide are added to the atmosphere per 3.8 kg of octane burned (hint: begin by writing a balanced equation for the combustion reaction.)
express your answer in kilograms to two significant figures.
mass =
kg
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Explanation:

Step1: Write the balanced chemical equation

The balanced chemical equation for the combustion of octane (\(C_{8}H_{18}\)) is \(2C_{8}H_{18}+25O_{2}
ightarrow16CO_{2}+18H_{2}O\)

Step2: Calculate the molar mass of octane and carbon dioxide

The molar mass of \(C_{8}H_{18}\): \(M_{C_{8}H_{18}}=(8\times12.01)+(18\times1.008)=96.08 + 18.144=114.224\space g/mol\)
The molar mass of \(CO_{2}\): \(M_{CO_{2}}=12.01+(2\times16)=44.01\space g/mol\)

Step3: Convert the mass of octane to moles

Given \(m_{C_{8}H_{18}} = 3.8\space kg=3800\space g\)
\(n_{C_{8}H_{18}}=\frac{m_{C_{8}H_{18}}}{M_{C_{8}H_{18}}}=\frac{3800}{114.224}\approx33.27\space mol\)

Step4: Use the mole - ratio from the balanced equation to find moles of \(CO_{2}\)

From the balanced equation \(2C_{8}H_{18}
ightarrow16CO_{2}\), the mole - ratio \(n_{CO_{2}}/n_{C_{8}H_{18}}=\frac{16}{2} = 8\)
\(n_{CO_{2}}=8\times n_{C_{8}H_{18}}\)
\(n_{CO_{2}}=8\times33.27 = 266.16\space mol\)

Step5: Convert moles of \(CO_{2}\) to mass

\(m_{CO_{2}}=n_{CO_{2}}\times M_{CO_{2}}\)
\(m_{CO_{2}}=266.16\times44.01\space g\)
\(m_{CO_{2}}=266.16\times44.01\times10^{- 3}\space kg\)
\(m_{CO_{2}}\approx11.71\space kg\)

Step6: Round to two significant figures

\(m_{CO_{2}}\approx12\space kg\)

Answer:

\(12\space kg\)