#2"Al"+3"S"→"Al"_2"S"_3# 125g of Al2S3 was produced from 75g of Al & an excess of S What is the theoretical yield of Al2S3?
1 Answer
Well, the theoretical yield is
You don't really need to know that
So, treat aluminum as your limiting reactant, the reactant whose consumption generates the maximum amount of product you can get.
#75 cancel"g Al" xx "1 mol"/(26.982 cancel"g Al") = "2.780 mols Al"#
And this many mols of
#"2 mols Al" = "1 mol Al"_2"S"_3#
So, this ratio is a unit conversion factor (i.e. a ratio that is equivalent to
#"2 mols Al"/("1 mol Al"_2"S"_3)#
And thus, we have...
#2.780 cancel"mols Al" xx ("1 mol Al"_2"S"_3)/(2 cancel"mols Al")#
#=# #"1.390 mols product"#
And now we just use its molar mass to find its produced mass:
#"1.390 mols Al"_2"S"_3 xx (2 xx "26.982 g" + 3 xx "32.065 g")/("1 mol Al"_2"S"_3)#
#=# #"208.69 g product"#
Unfortunately, we only get