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Question

# How many kilograms of $CO_2$ does the complete combustion of 3.8 kg of n-octane produce?

Solution

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The reaction

$\ce{C8H18(g)\;}+\;12.5 \ce{ \;O2(g) -> 8 CO2(g) + 9 H2O(g)}$

1 mol $n$-octane $\xrightarrow{produces}$ 8 mol $\ce{CO2}$

1 mol $\ce{CO2}$ = 44 g

1 mol $\ce{C8H18}$ = 114.23 g

Therefore, 114.23 g $n$-octane $\xrightarrow{produces}$ $8 \times 44$ g $\ce{CO2}$

Mass of $n$-octane = 3.8 kg = 3800 g

First, we convert mass of $\ce{C_8H_18}$ to moles using molar mass. Then, using mole ratio we convert moles of $\ce{C_8H_18}$ to moles of $\ce{CO_2}$. We find mass of $\ce{CO_2}$ gas produced after complete combustion of 3.8 kg $n$-octane using molar mass of $\ce{CO_2}$. Finally, convert the mass of $\ce{CO2}$ in g to kg.

\begin{aligned} \mathrm{3800\;g\;C_8H_{18} \times \dfrac{1\;mol\;C_8H_{18}}{114.23\;g\;C_8H_{18}} \times \dfrac{8\;mol\;CO_2}{1\;mol\;C_8H_{18}} \times \dfrac{44\;g\;CO_2}{1\;mol\;CO_2}}&\mathrm{\;= 11709.7\;g\;CO_2}\\ &=\mathrm{11.7\;kg\;CO_2} \end{aligned}

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