# Question 8d45c

Feb 2, 2018

Here's what I got.

#### Explanation:

Assuming that the question is correct, you can convert the number of moles of oxygen gas to grams by using its molar mass. Oxygen gas has a molar mass of ${\text{31.9988 g mol}}^{- 1}$, so you can say that $1$ mole of oxygen gas has a mass of $\text{31.9988 g}$.

This means that your reaction consumed

$11.52 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles O"_2))) * "31.9988 g"/(1color(red)(cancel(color(black)("mole O"_2)))) = color(darkgreen)(ul(color(black)("368.6 g}}}}$

The answer is rounded to four sig figs, the number of sig figs you have for the number of moles of oxygen gas that took part in the reaction.

$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

Seeing how this is supposed to be a stoichiometry problem, I think that you were actually looking for the mass of nonane that took part in the reaction.

You know by looking at the balanced chemical equation that the reaction consumes $9$ moles of oxygen gas for every $1$ mole of nonane that takes part in the reaction, so you can start by using this mole ratio to find the number of moles of nonane that reacted.

11.52 color(red)(cancel(color(black)("moles O"_2))) * ("1 mole C"_9"H"_20)/(9color(red)(cancel(color(black)("moles O"_2)))) = "1.28 moles C"_9"H"_20#

Now, to convert the number of moles of nonane to grams, use the compound's molar mass.

Nonane has a molar mass of ${\text{128.2 g mol}}^{- 1}$, which means that $1$ mole of nonane has a mass of $\text{128.2 g}$.

You can thus say that your reaction consumed

$1.28 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"_9"H"_20))) * "128.2 g"/(1color(red)(cancel(color(black)("mole C"_9"H"_20)))) = color(darkgreen)(ul(color(black)("164.1 g}}}}$

Once again, the answer must be rounded to four sig figs.