REACTION STOICHIOMETRY MONDAY SEPT 18 2006
“ICE CHARTS”
2 NaOH
+ H2SO4 
2 H2O
Na2SO4 +
INITIAL
CHANGE
- 2x
-x
+x
+2x
END
________________________________________________________________________
EXAMPLE ONE: THE REACTANTS ARE “STOICHIOMETRIC, FOLLOW THE COEFFICIENT
THEORETICAL MOLE RATIO EXACTLY.
OBJECTIVE: This example will show no unreacted reactant remains, total conversion of reactant
to product.
PROBLEM: If 0.5 mole of sulfuric acid is reacted with 1.0 moles of sodium hydroxide calculate
the moles of all species after the reaction has completed.
2 NaOH
+
H2SO4

Na2SO4
+
2 H2O
INITIAL
1.0 mol
0.50 mol
0.0 mol
0.0 mol
CHANGE
-2x
-x
+x
+2x
x
2x
END
result
1.0 mol - 2x 0.50 mol - x
( 0 mole)
(0 mol)
(0.50mol)
(1.0mol )
NOTE♫: LET X = 0.50. , when appropriate, set x to the species with a coefficient of one. In this
example x is 0.50.
EXAMPLE TWO: THERE IS A LIMITING REACTANT, ONE IS CONSUMED BEFORE THE OTHER,
THE LIMITING REACTANT IS COMPLETLEY CONSUMED AND SOME OF THE EXCESS
REACTANT PERSISTS TO CONTAMINATE THE PRODUCTS.
OBJECTIVE:
1. To identify if there is a limiting reagent
2. To identify the limiting reagent.
3. To predict yield with the limiting reagent by setting X in the ICE chart to it.
PROBLEM: If 1. mole of sulfuric acid is reacted with 1.5 moles of sodium hydroxide calculate the
moles of all species after the reaction has completed.
2 NaOH
+ H2SO4

Na2SO4
2 H2O
+
INITIAL
1.5 mol
1.0 mol
0.0 mol
0.0 mol
CHANGE
-2x
-x
+x
+2x
1.5 mol - 2x
1.0 mol - x
x
2x
END
1. To find the limiting reagent construct a ratio of the two reactants. For ease and consistency
place the one with the largest coefficient on top of the fractions. If the ACTUAL (mole
ratio from the problem or lab) is smaller that the THEORETICAL MOLE RATIO of
COEFFICIENTS, the top species is limiting.
Species
NaOH =
H2SO4
coefficient
Ratio
2
actual
ratio
=
1
1.5 mol
1.0 mol
In this example, the coefficient(theoretical) ratio is 2, the actual ratio is 1.5, and therefore NaOH
is the limiting reagent. Because the limiting reagent, NaOH is 1.5 moles, and it is 2x, therefore
X=0.75 mol.
Results:
NaOH = 1.5 – 2x ; 1.5 – 2(0.75) = 0 mol
H2SO4 = 1.0 – x ;
1.0 – 0.75
Na2SO4 = x = 0.75 mol
= 0.25 mol (unreacted)
H2O
= 2x = 1.5 mol