Question

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

Answer 1

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

`color(blue)(2)"Al"_ 2"O"_ (3(s)) + color(purple)(3)"C"_ ((s)) -> color(darkorange)(4)"Al"_ ((s)) + 3"CO"_ (2(g))`

Notice that the reaction consumes `color(blue)(2)` moles of aluminium oxide and `color(purple)(3)` moles of carbon and produces `color(darkorange)(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 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)("194 g"))) ->` the mass of carbon

`10.79 color(red)(cancel(color(black)("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"))) ->` 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.