Question #9ae68
1 Answer
Explanation:
Your starting point here will be the balanced chemical equation for this synthesis reaction
#color(red)(4)"Al"_text((s]) + 3"O"_text(2(g]) -> color(blue)(2)"Al"_2"O"_text(3(s])#
Notice that you have a
The problem tells you the oxygen gas is in excess, so you an assume the all the aluminium will react.
Use the molar mass of aluminium metal to determine how many moles you get in that
#4.6 color(red)(cancel(color(black)("g"))) * "1 mole Al"/(26.98color(red)(cancel(color(black)("g")))) = "0.1705"#
So, the theoretical yield of the reaction, which tells you how much product will be produced for a
#0.1705color(red)(cancel(color(black)("moles Al"))) * (color(blue)(1)color(white)(a)"moles Al"_2"O"_3)/(color(red)(2)color(red)(cancel(color(black)("moles Al")))) = "0.08525 moles Al"_2"O"_3#
Use aluminium oxide's molar mass to determine how many grams would contain that many moles
#0.08525color(red)(cancel(color(black)("moles Al"_2"O"_3))) * "101.96 g"/(1color(red)(cancel(color(black)("mole Al"_2"O"_3)))) = "8.69 g"#
So, if every mole of aluminium that takes part in the reaction ends up producing aluminium oxide, you can expect the reaction to form
However, you know that only
Percent yield is defined as
#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "what you actually get"/"what you should theoretical get" xx 100color(white)(a/a)|)))#
Well, a percent yield of
In this case, the actual yield of the reaction will be
#64color(red)(cancel(color(black)("%"))) * overbrace("8.69 g"/(100color(red)(cancel(color(black)("%")))))^(color(purple)("theoretical yield")) = "5.562 g"#
Rounded to two sig figs, the answer will be
#"actual yield" = color(green)(|bar(ul(color(white)(a/a)"5.6 g Al"_2"O"_3color(white)(a/a)|)))#