Transcript 3. Limiting Reactants

Warmup
1. Calculate the molar mass of SbI3
502.46 g/mole
2. According to the equation below, how many moles of
aluminum oxide can be made from 6.8 moles
aluminum?
4 Al
+ 3 O2
 2 Al2O3
6.8 moles Al
2
4
moles Al2O3 =
moles Al
3.4 moles Al2O3
a. What is Avogadro's
favorite kind of music?
b. What line from
Shakespeare do high school
moles have to memorize?
c. What element do moles
love to study in chemistry?
remolte
control
remoletly
sophmole
Rock 'N' Mole
"To mole or not to mole,
this is the question.“
Molybdenum
A device used by moles to watch Bill
Nye the Science Guy and the
Discovery Channel.
Obscurely having to do with a mole.
Anyone in the tenth grade who is
taking Chemistry already!
Limiting Reactants Notes
Conversion flow chart
Grams
given
Moles
given
Molar
mass
step
Moles
unknown
Mole-mole
conversion
step
Grams
unknown
Molar
mass
step
Problem: If you start with X grams of Givenium, how
many grams of Unknownium will be made?
X grams of given (1 mole given )( B moles unknown)(D grams of unknown) =
(A grams of given) ( C moles given ) ( 1 mole unknown )
Key:
Given = the compound whose amount is given to you in the problem
Unknown = compound whose amount you are looking for
X = grams that the problem gives you
A = molar mass of the compound given in the problem (see periodic table)
B and C = numbers from molar ratio, found in the chemical equation
D = molar mass of the unknown compound from the problem (see periodic table)
Ex 1: How many grams of Al2O3 can be formed from
5.81 grams of Al?
Al + 3 O2
 2 Al2O3
4
1. Write/Balance Equation:
2. Figure out Given: 5.81 grams Al
3. Choose correct pathway:
Figure out Unknown: ? grams Al2O3
g Al  mol Al  mol Al2O3  g Al2O3
4. Numbers!
5.81 g Al (
1
(
mol Al) ( 2
26.98 g Al) (
mol Al2O3 )( 101.96 g Al2O3 )
4
mol Al)
(
1
= 11.0 grams Al2O3
mol Al2O3 )
Ex 2: Hydrogen gas and chlorine gas react to form hydrochloric
acid (HCl). How many grams of the product will form in the
reaction if 3.75 grams of chlorine gas is used?
1 H
+ 1 Cl2  2 HCl
2
1. Write/Balance Equation:
2. Figure out Given: 3.75 grams Cl2
3. Choose correct pathway:
Figure out Unknown: ? grams HCl
g Cl2  mol Cl2  mol HCl  g HCl
4. Numbers!
3.75
g Cl2
(
(
1
70.9
mol Cl2 )
g Cl2 ) (
(
2
1
mol HCl )
mol Cl2 )
(
36.46
(
1
g HCl)
mol HCl )
= 3.86 grams HCl
2Sb(s) + 3I2(s) → 2SbI3(s
Ex 3. What mass of antimony(III) iodide can form
from 1.20g Sb ?
1.20 g Sb
(
(
1
mol Sb )
121.76 g Sb ) (
(
2
mol SbI3)
(
502.46
2
mol Sb)
(
1
g SbI3 )
mol SbI3 )
= 4.94 g SbI3
Wait! We know how much Sb we have. But what if
there is just a tiny, tiny amount of iodine available?
Hmmmm…
Limiting Reactant – reactant that is used up first in a
reaction; it limits how much product you can make!!!
2Sb(s) + 3I2(s) → 2SbI3(s)
Ex 4. Determine the theoretical yield (mass in
grams) of antimony(III) iodide formed when 1.20g
Sb and 2.40 g I2 are mixed.
2.40 g I2
(
(
1
253.8
= 3.17 g SbI3
mol I2 )
(
2
mol SbI3)
(
502.46
g I2 )
(
3
mol I2)
(
1
g SbI3 )
mol SbI3 )
wait but 1.20 g Sb predicts 4.94 g SbI3
5. Do another calculation with the other amount given
6. Choose the lower amount…the higher amount is not possible. The
theoretical yield is calculated based upon the limiting reactant.
3.17 g SbI3. I2 is the limiting reactant (even though there is a greater
amount of it) and is completely used up. Some Sb will be leftover.
Ex 5. Suppose 36.0 grams of ammonia and 50.0 grams of
oxygen react to form nitrogen gas and water. Calculate
the mass of nitrogen gas that could be formed.
4 NH3 + 3 O2  2
N2 + 6 H2O
36.0 g NH3 (
1
mol NH3 )
( 17.04
(
g NH3) (
2
mol N2)
(
28.02
4
mol NH3)
(
1
=
50.0 g O2
(
(
1
32.00
mol O2 )
g O2 ) (
(
2
3
mol N2)
mol O2)
(
g N2 )
mol N2 )
29.6 g N2
28.02
( 1
g N2 )
mol N2 )
= 29.2 g N2
Ex 6. From the previous problem, what mass of excess
reactant remains?
4 NH3 + 3 O2  2
N2 + 6 H2O
29.2 g N2 were produced. Oxygen gas = limiting
reactant. Rephrase this question: how much
ammonia is required to produce 29.2 g N2 ?
29.2 g N2
(
(
1
28.02
mol N2 )
g N2 ) (
(
4
2
mol NH3) (
17.04
(
mol N2)
=
1
g NH3 )
mol NH3 )
35.5 g NH3
35.5 g were actually required but 36.0 grams NH3 were used in
the reaction. Therefore, the excess NH3 = 0.5 g
Ex 7. In the previous problem, we calculated
that 29.2 g N2 should be produced. If 26.7
grams N2 are actually produced, calculate
the percent yield of the experiment
percent yield = actual yield x 100
theoretical yield
percent yield = 26.7 g x 100
29.2 g
91.4 %
Conversion flow chart
#molecules /atoms
/ions given
Grams
given
# unknown molecules /
atoms/ions
Moles
given
Molar
mass
step
Moles
unknown
Mole-mole
conversion
step
Grams
unknown
Molar
mass
step
Ex 8: 6.7 x 1017 molecules chlorine gas can form what
mass of hydrochloric acid? Assume an excess of
hydrogen gas.
1
H2
+
Pathway:
1
Cl2
2

HCl
molec. Cl2  mol Cl2  mol HCl  g HCl
6.7 x 1017 molec Cl2 (
(
6.02x 1023
1
mol Cl2 ) (
molec. Cl2 ) (
2
mol HCl ) (
1
mol Cl2 ) (
36.46
1
g HCl)
mol HCl )
= 8.1 x 10-5 grams HCl
Ex 9: 98.7 g Cl2 will completely react with how
many moles of hydrogen gas?
1
H2
+
1

Cl2
Pathway:
2
HCl
g Cl2  mol Cl2  mol H2
98.7 g Cl2
(
(
70.9
1
mol Cl2 ) ( 1
g Cl2 )
(
mol H2 )
1
mol Cl2 )
= 1.39 moles H2