REDOX STOICHIOMETRY
1.
For the reaction...
4 Zn + 10 H 1 +(aq) + NO 3 1 -(aq) =
4 Zn 2 +(aq) + NH 4 1+(aq) + 3 H 2 O
What mass of zinc is required to react
completely with 22.54 mL of a 0.153 M
solution of aluminum nitrate?
2.
For the reaction...
3 OH 1 - + 6 V
+ 14 H 2 O
HV 6 O17 3 -
=
+ 15 H 2
What mass of hydrogen gas will be produced
from reaction of 45.00 mL of a 0.095 M
solution of barium hydroxide with 0.775 grams
of vanadium?
1.
Nitrate anion is a reactant in this reaction. However, nitrate anion cannot exist as an
independent entity, some counterbalancing cation must always be associated with the
anion. In this case, a solution of aluminum nitrate furnishes the nitrate anion to the
reaction, and the concentration of the solution is expressed in terms of Al(NO 3 ) 3 .
So...
moles zinc = moles aluminum nitrate [ C.F.'s]
moles of ZINC (a solid substance) =
moles aluminum nitrate (a solution)
then:
 given mass   ? grams 

=
 std . mass   65.4 g / mole 
=
( Molarity )(Vol. Liters)
(? grams / 65.4) = ( 0.153 )(0.02254 L) [ C.F.'s]
 3 mole NO31−  4 mole Zn 
[ C.F.'s] = 

1− 
1 mole Al ( NO3) 3 1 mole NO3 
The FIRST C.F. is based on the chemical formula of aluminum nitrate, and relates
ONE mole of compound aluminum nitrate, to THREE moles of nitrate anions.
The SECOND C.F. is based on coefficients in the balanced chemical equation, and
relates FOUR moles of zinc to ONE mole of nitrate anion.
? grams = 2.7065 grams zinc
2.
Recognize this is a limiting reagent type of a problem, because amounts of both
reactants are specified.
The barium hydroxide solution furnishes hydroxide anions to the reaction. Since the
molarity and volume of the solution are both given, and also because the formula of
barium hydroxide is known, then the amount (i.e., moles) of hydroxide anion available
to the reaction is also given.
Similarly, because the mass of solid vanadium is given, and its standard mass is
known, then the amount (i.e., moles) of vanadium is also given.
Solution to this problem will be illustrated using the B4, REACT, and AFTR grid.
Let's use millimole units, and calculate millimole quantities of hydroxide and vanadium:
millimoles = ( Molarity ) ( Vol. in mL )
millimoles = ( mass in mg / Std.mass )
? millimoles OH −
= millimoles Ba(OH) 2 [ C.F. ]
 2 mole OH − 
= (0.095 M)(45.00) 
= 8.55 mmoles OH −

1 mole Ba ( OH ) 2 
? millimoles V
= 775 milligrams V
 1 millimole 
50.94 milligrams  = 15.21 mmoles V


set-up the reaction grid, and enter these starting amounts in the B4 row...
REACTANTS, will be lost
B4
REACT, in ratio of
( -3; -6; +15 )
AFTR
PRODUCTS
15 H 2
3 OH − 6 V
8.55 15.21
none
-7.61 -15.21 +38.025
0.94 none 38.025
Suppose millimoles hydroxide was picked as a starting point. How much vanadium
would be needed to completely react with 8.55 millimoles of hydroxide anion? Let's
figure it out...
? millimoles V = 8.55 millimoles OH − [ C.F. ]
The relationship between V and OH − is given by coefficients in the balanced chemical
redox equation; 6 V for every 3 OH −. So the C.F. is as shown...
? millimoles V = 8.55 millimoles OH −
 6 millimoles V 
 3 millimoles OH − 
= 17.1
Thus 17.1 millimoles V is needed to react with all of the hydroxide anion present.
However, only 15.21 millimoles V is actually present, so hydroxide anion must be the
reagent present in an excess amount - or vanadium is the limiting reagent.
Conclude that all vanadium will be used up in this reaction. The amount of hydroxide
actually needed for this reaction will then be half the amount of vanadium reacted,
because their coefficients are in the ratio 3 OH − for every 6 V.
( 3 / 6 ) of 15.21 = 7.61 millimoles OH −
The amount of hydrogen formed will be more than the amount of vanadium according
to the ratio 15 to 6, because their coefficients are in the ratio 15 H 2 for every 6 V.
( 15 / 6 ) of 15.21 = 38.025 millimoles H 2
Be certain that ALL quantities present in the REACT ROW are in the same proportion
as coefficients in the balanced chemical equation.
In this instance the proportions (via coefficients) are -3 OH, to -6 V, to +15 H 2 ,
are in the same proportions as (via millimole amounts)
-7.61 millimoles OH −, to -15.21 millimoles V, to +38.025 millimoles H 2 .
Finallly, what is the mass of 38.025 millimoles of hydrogen?
millimoles =
 mass in milligrams


 standard mass 
38.025 millimoles = ? milligrams / 2
? milligrams H 2
= 76.05
or,
? grams H 2
= 0.076