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

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1 Answer
Mar 28, 2017

Answer:

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.