Download Stoichiometric Problems III: Sto c o et c ob e s

Stoichiometric
Sto
c o et c Problems
ob e s III:
Limiting reactants
In a limiting reactant problem you are given
either g
gram or mole amounts of starting
g
reactants, and asked to determine which
reactant runs out first, and ‘limits’ the
amount of product that can be made
made.
I have found that the method that the book
uses for
f these
th
problems
bl
tends
t d to
t confuse
f
people, so, I will use a different method that
follows directly what you have learned so far.
Stoichiometric
Sto
c o et c Problems
ob e s III:
Limiting reactants
Here is a typical problem.
problem If I have the
following chemical reaction:
N2(g) + 3H2 (g) = 2NH3 (g)
And I start with 5 moles of N2 and 5 moles of
H2, which chemical (H2 or N2) is the limiting
reactant?
My approach to solving this kind of problem
is to calculate the amount of product that
would be formed from each reagent. The
reagent that makes the smallest amount of
product is your limiting reactant because it
limits the overall amount of the reaction.
N2(g) + 3H2 (g) = 2NH3 (g)
If I start
t t with
ith 5 moles
l off N2 and
d 5 moles
l off H2,
which chemical (H2 or N2) is the limiting reactant?
So your job is to calculate the moles of
product from each reactant:
N2:
H2:
5 mole N2 x (2 mole NH3/1 mole N2) =
10 moles NH3
5 mole H2 x (2 mole NH3/3 mole H2) =
3.3 mole NH3
You can see that the 5 moles of H2 makes
much less product, so if we started with 5
moles of each reactant,
reactant this one would run
out first and limit the reaction to 3.3 moles
of product.
Stoichiometric
Sto
c o et c Problems
ob e s III:
Limiting reactants
Now let’s make it a little more interesting;
l ’ start with
let’s
i h grams off reactants instead
i
d off
moles, and I’ll give you 2 products instead
on one.
Given the balanced chemical reaction:
C3H8(g) + 5O2(g) = 3CO2(g) + 4 H2O(g)
If you start with 5 grams of C3H8(g) and 10
grams of O2(g), which is the limiting
reactant?
t t?
In the reaction: C3H8(g) + 5O2(g) = 3CO2(g) + 4
H2O(g)
O( ) If you start
t t with
ith 5 grams off C3H8(g)
( ) and
d
10 grams of O2(g), which is the limiting reactant?
The last complication, which of the two
products
d
you want to use, is
i really
ll no
problem, choose whichever one you want.
It doesn’t matter,, just
j
be sure that y
you don’t
change products in the middle of the
problem.
I will calculate the moles of CO2 product,
simply because it is the first product in the
equation.
ti
In the reaction: C3H8(g) + 5O2(g) = 3CO2(g) + 4
H2O(g)
O( ) If you start
t t with
ith 5 grams off C3H8(g)
( ) and
d
10 grams of O2(g), which is the limiting reactant?
How do you handle the fact that I started
you iin grams off reactants instead
i
d off moles?
l ?
Again
g
a simple
p answer,, convert to moles and
you are on your way.
1 mole of C3H8 = (3(12
(3(12.01)
01) + 8(1
8(1.008))
008)) =
=80.124g
1 mole O2 = 2(16.00) = 32.00g
In the reaction: C3H8(g) + 5O2(g) = 3CO2(g) + 4
H2O(g)
O( ) If you start
t t with
ith 5 grams off C3H8(g)
( ) and
d
10 grams of O2(g), which is the limiting reactant?
C3H8:
1 mole C3 H8
3 mole CO2
5g C3 H8 ×
×
= .19 mole CO2
. g C3 H8 1 mole C3 H8
80124
O2:
1 mole O2
3 mole CO2
5g O2 ×
×
= .09 mole CO2
32.00 g O2
5 mole O2
In the reaction: C3H8(g) + 5O2(g) = 3CO2(g) + 4
H2O(g)
O( ) If you start
t t with
ith 5 grams off C3H8(g)
( ) and
d
10 grams of O2(g), which is the limiting reactant?
So in this problem the O2 was the limiting
reactant because
b
it
i only
l made
d .09
09 mole
l off
CO2 product.
Stoichiometric
Sto
c o et c Problems
ob e s III:
Limiting reactants
Okay, here is a practice problem for you.
If I start the reaction 2CuO(s) + C(s) =
2Cu(s)
( ) + CO2(g) with exactly
y .5 g of CuO
and 1 g of C, which reactant, C or CuO is the
limiting reactant?
(Answer is on the next slide, work it yourself
and then look at the answer.)
If I start the reaction 2CuO(s) + C(s) = 2Cu(s) +
CO2(g)
( ) with
ith exactly
tl .5
5 g off C
CuO
O and
d 1 g off C,
C which
hi h
reactant, C or CuO is the limiting reactant?
1 mole CuO
2 mole Cu
.5g CuO ×
×
= .006moleCu
79.55g CuO 2 mole CuO
1 mole C 2 mole Cu
1g C ×
×
= .17 mole
l Cu
C
12.01g C 1 mole C
.5g CuO is the limiting reactant