# How much aluminum oxide and how much carbon are needed to prepare 454 g of aluminum by the balance equation? See picture!

## Mar 28, 2017

Here's what I got.

#### Explanation:

The first thing to do here is to convert the mass of aluminium to moles by using the element's molar mass

454 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "16.83 moles Al"

Now, you know that the chemical equation that describes this reaction looks like this

$\textcolor{b l u e}{2} {\text{Al"_ 2"O"_ (3(s)) + color(purple)(3)"C"_ ((s)) -> color(darkorange)(4)"Al"_ ((s)) + 3"CO}}_{2 \left(g\right)}$

Notice that the reaction consumes $\textcolor{b l u e}{2}$ moles of aluminium oxide and $\textcolor{p u r p \le}{3}$ moles of carbon and produces $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{4}$ moles of aluminium.

This represents the reaction's theoretical yield, i.e. what you get if the reaction has a 100% yield.

You can thus say that for a 100% yield, you would need

16.83 color(red)(cancel(color(black)("moles Al"))) * (color(purple)(3)color(white)(.)"moles C")/(color(darkorange)(4)color(red)(cancel(color(black)("moles Al")))) = "12.62 moles C"

and

16.83 color(red)(cancel(color(black)("moles Al"))) * (color(blue)(2)color(white)(.)"moles Al"_ 2"O"_ 3)/(color(darkorange)(4)color(red)(cancel(color(black)("moles Al")))) = "8.415 moles Al"_ 2"O"_ 3

in order to produce $16.83$ moles of aluminium. However, you know that the reaction has a 78% yield, which implies that for every $100$ moles of aluminium that could be produced, the reaction ends up producing $78$ moles.

You can thus say that in order for the reaction to have an actual yield of $16.83$ moles, it must have a theoretical yield of

16.83 color(red)(cancel(color(black)("moles Al"))) * "100 moles Al needed"/(78color(red)(cancel(color(black)("moles Al produced")))) = "21.58 moles Al"

You can now use the same mole ratios to determine how much carbon and aluminium oxide are needed in order to account for a theoretical yield of $21.58$ moles of aluminium, which will be equivalent to an actual yield of $16.83$ moles.

21.58 color(red)(cancel(color(black)("moles Al"))) * (color(purple)(3)color(white)(.)"moles C")/(color(darkorange)(4)color(red)(cancel(color(black)("moles Al")))) = "16.19 moles C"

and

21.58 color(red)(cancel(color(black)("moles Al"))) * (color(blue)(2)color(white)(.)"moles Al"_ 2"O"_ 3)/(color(darkorange)(4)color(red)(cancel(color(black)("moles Al")))) = "10.79 moles Al"_ 2"O"_ 3

To convert these values to grams, use the molar masses of the two reactants

$16.19 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)("194 g}}}} \to$ the mass of carbon

$10.79 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles Al"_ 2"O"_ 3))) * "101.96 g"/(1color(red)(cancel(color(black)("mole Al"_ 2"O"_ 3)))) = color(darkgreen)(ul(color(black)(1.10 * 10^3color(white)(.)"g}}}} \to$ the mass of aluminium oxide

I'll leave the answers rounded to three sig figs, but don't forget that you only have two sig figs for the percent yield.