# Consider the unbalanced equation: C_6H_14 + O_2 -> CO_2 + H_2O. What mass of O_2 is required to react with "11.5 g" of C_6H_14?

Mar 15, 2017

$\text{0.0400 g O"_2}$ is required to react with $\text{11.5 g}$ of $\text{C"_6"H"_14}$.

#### Explanation:

${\text{2C"_6"H"_14 + "19O}}_{2}$$\rightarrow$$\text{12CO"_2 + "14H"_2"O}$

Use the balanced equation to determine the mole ratios between $\text{C"_6"H"_14}$ and $\text{O"_2}$.

$\left(\text{2 mol C"_6"H"_14)/("19 mol O"_2}\right)$ and $\left({\text{19 mol O"_2)/("2 mol C"_6"H}}_{14}\right)$

Determine the molar masses of ${\text{C"_6"H}}_{14}$ and $\text{O"_2}$.

${\text{C"_6"H}}_{14} :$$\text{86.178 g/mol}$
https://www.ncbi.nlm.nih.gov/pccompound?term=C6H14
${\text{O}}_{2} :$$\text{31.998 g/mol}$

Determine the moles of $\text{C"_6"H"_14}$ by dividing its given mass by its molar mass.

(11.5 cancel"g C"_6"H"_14)/(86.178cancel"g"/"mol")="0.13344 mol C"_6"H"_14"

Determine moles of $\text{O"_2}$ by multiplying mol $\text{C"_6"H"_14}$ by the mol ratio that has $\text{O"_2}$ in the numerator.

$0.13344 \cancel{\text{mol C"_6"H"_14xx(19"mol O"_2)/(2 cancel"mol C"_6"H"_14)="1.2678 mol O"_2}}$

Determine the mass of $\text{O"_2}$ that will react with $11.5 \text{g C"_6"H"_14}$ by multiplying the moles $\text{O"_2}$ by its molar mass.

$1.2678 \cancel{\text{mol O"_2xx(31.998"g O"_2)/(1cancel"mol O"_2)="0.0400 g O"_2}}$ (rounded to three significant figures)