5)Sulfur and oxygen react in a combination reaction to produce sulfur trioxide, an
environmental 5)
pollutant:
2S (s) + 3O2 (g) 2SO3 (g)
In a particular experiment, the reaction of 1.0 g S with 1.0 g O2 produced 0.80 g of SO3.
The %
yield in this experiment is __________.
AND calculate the moles of unreacted excess reagent.
Calculate the moles of the excess reagent that is unconsumed at the end ofreaction.
MOLES = MASS/GFM S
MOLES = 1.0 g s/ 32.04g/mol
MOLES = 0.031191 mol S
MOLES = MASS/GFM O2
MOLES = 1.0 g O2 / 32.00g/mol
MOLES = 0.03125mol O2
S
O2
= 2 = 0.031191 ,
3
0.03125
.666 <
1
The actual ratio is larger than the
theoretical, therefore the actual ratio
fraction is too large, the denominator is
to small,
O2 limiting.
INTRODUCING ICE CHART SOLUTION SKILL
THE ICE CHART WILL ALLOW YOU TO CALCULATE MOLES OF ALL
SPECIES AFTER THE REACTION IS COMPLETED.
MOLES OF ALL PRODUCTS.
MOLES OF ALL EXCESS REACTANTS REMAINNING.
THE FASTEST WAY TO APPROACH A REACTION IN WHICH YOU
ARE CALCULATING MANY SUBSTANCES AT ONCE.
2S
INITIAL
CHANGE
END
+
0.031191 mol
-2X
0.031191 – 2X
3O2

2SO3
0.03125mol
LIMITING
0.0
-3X
-3X
0.0
3X
DETERMINING THE VALUE OF “X”
1. X IS DETERMINED BY THE LIMITING REAGENT.
2. IN THIS PROBLEM 0.03125mol = 3X FROM
THE LIMITING REAGENT OXYGEN. X = 0.0104111
MOL.
3. THE EXCESS REAGENT IS SULFER, WHICH IS
0.031191 – 2X
0.031191 – 2(0.0104111)
0.01036 MOLE OF S REMAINS
4. THE PRODUCT SO3 = 2X = 2(0.0104111) =
0.0208222 MOL ( 1.667 GRAMS)