Question #1f048
1 Answer
Explanation:
In order to find the percent yield of a given reaction, you need to know two things
- the theoretical yield of the reaction, i.e .how much product you would get if the reaction had a
#100%# yield- the actual yield of the reaction, i.e. how much product you actually get
In your case, you know that the reaction produced
In order to find the theoretical yield, you need to use the mole ratio that exists between iron(III) oxide,
The balanced chemical equation for this reaction looks like this
#"Fe"_2"O"_text(3(s]) + 2"Al"_text((s]) -> "Al"_2"O"_text(3(s]) + color(red)(2)"Fe"_text((s])#
Notice that every mole of iron(III) oxide that takes part in the reaction will produce
Since aluminium is said to be in excess, you can assume that all the moles of iron(III) oxide will take part in the reaction. Use the compound's molar mass to determine how many moles you have in that
#30.00color(red)(cancel(color(black)("g"))) * ("1 mole Fe"_2"O"_3)/(159.69color(red)(cancel(color(black)("g")))) = "0.18786 moles Fe"_2"O"_3#
So, if this is how many moles of iron(III) oxide take part in the reaction, you can use the aforementioned
#0.18786color(red)(cancel(color(black)("moles Fe"_2"O"_3))) * overbrace((color(red)(2)color(white)(a)"moles Fe")/(1color(red)(cancel(color(black)("mole Fe"_2"O"_3)))))^(color(purple)("1:2 mole ratio")) = "0.37572 moles Fe"#
This many moles would correspond to a mass of
#0.37572color(red)(cancel(color(black)("moles Fe"))) * overbrace("55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))))^(color(brown)("molar mass of Fe")) = "20.982 g Fe"#
So, the reaction should theoretically produce
#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" xx 100color(white)(a/a)|)))#
Plug in your values to get
#"% yield" = (18.25 color(red)(cancel(color(black)("g"))))/(20.982color(red)(cancel(color(black)("g")))) xx 100 = color(green)(|bar(ul(color(white)(a/a)"86.98%"color(white)(a/a)|)))#
The answer is rounded to four sig figs.