Chemistry: 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.

1.What mass of oxygen gas,O2, from the air is

Ответ:

gungamer720

27.01.2022

Chemistry

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:

$\rm C$ :

. $\rm H$ : 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

, whileThe coefficient of $\rm C_8 H_{18}$ is

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 $\rm O_2$ ?

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

0,0(0 оценок)

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.

Ответ:

Ответ:

Ask your question to the AI advisor

More tips

Answers on questions: Chemistry

Ask an AI advisor a question