How much aluminum oxide and how much carbon are needed to prepare 454 g of aluminum by the balance equation? See picture!
1 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
This represents the reaction's theoretical yield, i.e. what you get if the reaction has a
You can thus say that for a
#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
You can thus say that in order for the reaction to have an actual 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 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.