Question

1.What mass of oxygen gas,O2, from the air is consumed in the combustion of 702g of octane, C8H18, one of the principal components of gasoline? 2C8H18 + 25O2 -> 16CO2 + 18H2O. ​

Answer 1

Answer:

Approximately $2.46\times 10^3\; \rm g$.

Explanation:

Molar mass of C₈H₁₈ and O₂

Look up the following relative atomic mass values on a modern periodic table:

• $\rm C$: $12.011$.
• $\rm H$: $1.008$.
• $\rm O$: $15.999$.

Calculate the molar mass of octane, $\rm C_8 H_{18}$:

$M(\rm C_8 H_{18}) = 8 \times 12.011 + 18 \times 1.008 = 114.232\; \rm g \cdot mol^{-1}$.

Similarly, calculate the molar mass of $\rm O_2$:

$M(\mathrm{O_2}) = 2 \times 15.999 = 31.998 \; \rm g \cdot mol^{-1}$.

Number of moles of octane molecules

How many moles of molecules in $702\; \rm g$ of octane, $\rm C_8 H_{18}$?

$\displaystyle n({\rm C_8 H_{18}}) = \frac{m({\rm C_8 H_{18}})}{M({\rm C_8 H_{18}})} = \frac{702\; \rm g}{114.232\; \rm g \cdot mol^{-1}}\approx 6.1454\; \rm mol$.

Number of moles of O₂ required

Make sure that the chemical equation here is balanced.

In this equation,

• The coefficient of $\rm O_2$ is $25$, while
• The coefficient of $\rm C_8 H_{18}$ is $2$.

Ratio between the two coefficients:

$\displaystyle \frac{\text{Coefficient of \mathrm{O_2} }}{\text{Coefficient of \mathrm{C_8H_{18}} }} = \frac{25}{2}$.

Therefore:

$\displaystyle \frac{n(\mathrm{O_2},\, \text{consumed})}{n(\mathrm{C_8H_{18}},\, \text{consumed})} = \frac{25}{2}$.

Rearrange to obtain:

$\displaystyle n(\mathrm{O_2},\, \text{consumed}) = \frac{25}{2}\times n(\mathrm{C_8H_{18}},\, \text{consumed})$.

It is already determined that $\displaystyle n({\rm C_8 H_{18}})\approx 6.1454\; \rm mol$. Hence, the number of moles of oxygen gas required will be:

$\begin{aligned}& \frac{25}{2}\times n(\mathrm{C_8H_{18}}) \approx \frac{25}{2}\times 6.1454\; \rm mol \approx 76.817\; \rm mol\end{aligned}$.

Mass of O₂ required

What would be the mass of that $76.817\; \rm mol$ of $\rm O_2$?

$\begin{aligned}&m({\rm O_2}) \\ &= n({\rm O_2}) \cdot M({\rm O_2})\\ &\approx 76.817\; \rm mol \times 31.998\; \rm g \cdot mol^{-1} \\ &\approx 2.46\times 10^3\; \rm g\end{aligned}$.