Download Worksheet 3 Answers

Limiting Reactants
Adjusting to Reality
Unit 8 WS 3
1 N2 + 3 H2 
BEFORE
2 NH3
AFTER
How many NH3 molecules are produced? 8
Which reactant is in excess & why? N2
It’s left over
How many excess molecules are there? 2
Constructing a BCA Table
1 N2
Equation
Before
6 mol
+
3 H2
12 mol

2 NH3
0 mol
Change
After
Let’s say you don’t see right away which reactant will get
used up…So we’ll just take a guess that the Nitrogen
gets used up.
Constructing a BCA Table
1 N2
Equation
+
3 H2
Before
6 mol
12 mol
Change
- 6 mol
- 18 mol
0 mol
WRONG
After

2 NH3
0 mol
Let’s say you don’t see right away which reactant will get
used up…So we’ll just take a guess that the Nitrogen
gets used up.
Constructing a BCA Table
1 N2
Equation
+
3 H2

2 NH3
Before
6 mol
12 mol
0 mol
Change
- 4 mol
-12 mol
+ 8 mol
2 mol
0 mol
8 mol
After
Now we know the hydrogen has to get used up.
Constructing a BCA Table
1 N2
Equation
+
3 H2

2 NH3
Before
6 mol
12 mol
0 mol
Change
- 4 mol
-12 mol
+ 8 mol
2 mol
0 mol
8 mol
After
How many NH3 can be produced? 8
Which reactant is in excess & why? N2
There’s some in the after line
How many molecules of N2 are there left? 2
What must you look for in a particular
reactant mixture to decide what
reactant will be in excess?
Both the coefficients in the balanced equation
& how much of each reactant you start with.
# 3. When 0.50 mol of Al reacts with 0.72 mol
of I2, how many moles of the excess reactant
will remain? How many moles of aluminum
iodide will be formed?
Equation
Before
Change
After
2 Al
+
3 I2

2 Al I3
# 3. When 0.50 mol of Al reacts with 0.72 mol
of I2, how many moles of the excess reactant
will remain? How many moles of aluminum
iodide will be formed?
Equation
Before
Change
After
2 Al
.50 mol
+
3 I2
.72 mol

2 Al I3
# 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of
the excess reactant will remain? How many moles of aluminum iodide will
be formed?
Equation
2 Al
+
3 I2

2 Al I3
Before
.50 mol
.72 mol
0 mol
Change
-.48 mol
-.72 mol
+.48 mol
After
.02 mol
0 mol
.48 mol
Looking at these numbers, some of us might see that if there’s a 2 in front of
the Al and I have .50 moles of it, I would need .75 moles of the I2 to react
with it, & since I have less than that, I know it’s the I2 that runs out.
If you don’t see this – you can always guess & test.
# 3. When 0.50 mol of Al reacts with 0.72 mol of I2, how many moles of
the excess reactant will remain? How many moles of aluminum iodide will
be formed?
Equation
2 Al
+
3 I2

2 Al I3
Before
.50 mol
.72 mol
0 mol
Change
-.48 mol
-.72 mol
+.48 mol
After
.02 mol
0 mol
.48 mol
Moles of excess reactant: 0.02 mol Al
Moles of AlI3 formed: 0.48 mol AlI3
# 4. When sodium hydroxide reacts with sulfuric acid
(H2SO4), water and sodium sulfate are the products.
Calculate the mass of sodium sulfate produced when 15.5 g of
sodium hydroxide are reacted with 46.7 g of sulfuric acid.
Equation
Before
Change
After
2 NaOH +
1
H2SO4

2 H2O
+
Na2SO4
# 4. When sodium hydroxide reacts with sulfuric acid
(H2SO4), water and sodium sulfate are the products.
Calculate the mass of sodium sulfate produced when 15.5 g of
sodium hydroxide are reacted with 46.7 g of sulfuric acid.
1 mol NaOH
15.5 g NaOH
x
= 0.388 mol NaOH
40.00 g NaOH
1 mol H2SO4
46.7 g H2SO4
x
= 0.473 mol H2SO4
98.72 g H2SO4
# 4. When sodium hydroxide reacts with sulfuric acid
(H2SO4), water and sodium sulfate are the products.
Calculate the mass of sodium sulfate produced when 15.5 g of
sodium hydroxide are reacted with 46.7 g of sulfuric acid.
Equation
Before
Change
After
2 NaOH +
.388 mol
1
H2SO4
.473 mol

2 H2O
0 mol
+
Na2SO4
0 mol
# 4. When sodium hydroxide reacts with sulfuric acid
(H2SO4), water and sodium sulfate are the products.
Calculate the mass of sodium sulfate produced when 15.5 g of
sodium hydroxide are reacted with 46.7 g of sulfuric acid.
Equation
2 NaOH +
1
H2SO4

2 H2O
+ Na2SO4
Before
.388 mol
.473 mol
0 mol
0 mol
Change
-.388 mol
-.194 mol
+.388 mol
+.194 mol
0 mol
.279 mol
.388 mol
.194 mol
After
Looking at these numbers, some of us might see that if there’s a 2 in front of
the NaOH and I have .388 moles of it, I would need only half of that (about
1.9 off the top of my head) for the sulfuric acid, so I must use up the NaOH.
If you don’t see this – you can always guess & test.
# 4. Calculate the mass of sodium sulfate produced when 15.5
g of sodium hydroxide are reacted with 46.7 g of sulfuric acid.
142.1 g Na2SO4
0.194 mol Na2SO4
x
= 27.6 g Na2SO4
1 mol Na2SO4
# 5. A 14.6 g sample of oxygen gas is placed in a sealed
container with 2.5 g of hydrogen gas. The mixture is
sparked, producing water vapor. Calculate the mass of
water formed. Calculate the number of moles of the excess
reactant remaining.
Equation
Before
Change
After
2 H2
+
1 O2

2 H2O
# 5. A 14.6 g sample of oxygen gas is placed in a sealed
container with 2.5 g of hydrogen gas. The mixture is sparked,
producing water vapor. Calculate the mass of water formed.
Calculate the number of moles of the excess reactant
remaining.
1 mol O2
14.6 g O2
x
= 0.456 mol O2
32.0 g O2
1 mol H2
2.5 g H2 x
= 1.24 mol H2
2.02 g H2
# 5. A 14.6 g sample of oxygen gas is placed in a sealed
container with 2.5 g of hydrogen gas. The mixture is
sparked, producing water vapor. Calculate the mass of
water formed. Calculate the number of moles of the excess
reactant remaining.
Equation
Before
Change
After
2 H2
1.24 mol
+
1 O2
0.456 mol

2 H2O
0 mol
# 5. A 14.6 g sample of oxygen gas is placed in a sealed
container with 2.5 g of hydrogen gas. The mixture is
sparked, producing water vapor. Calculate the mass of
water formed. Calculate the number of moles of the excess
reactant remaining.
Equation
2 H2
+
1 O2

2 H2O
Before
1.24 mol
0.456 mol
0 mol
Change
-.912 mol
- .456 mol
+ .912 mol
.33 mol
0 mol
.912 mol
After
Looking at these numbers, some of us might see that if there’s a 1 in front of
the O2 and I have .456 moles of it, I would need double that (about 0.9 off
the top of my head) for the H2, so I must use up the O2.
If you don’t see this – you can always guess & test.
# 5. Calculate the mass of water formed. Calculate the
number of moles of the excess reactant remaining.
18.02 g H2O
0.912 mol H2O
x
= 16.4 g H2O
1 mol H2O
# 6. How much PEA can be made from 75.0g of ammonium
formate and 125g of acetophenone? What mass of the
excess reactant remains?
Equation
Before
Change
After
CH5NO2
+
C8H8O 
C8H11N
+
CO2
+
H2O
# 6. How much PEA can be made from 75.0g of ammonium
formate and 125g of acetophenone? What mass of the
excess reactant remains?
1 mol AF
75.0 g AF
x
= 1.19 mol AF
63.07 g AF
1 mol ACE
125 g ACE x
= 1.04 mol ACE
120.16 g ACE
# 6. How much PEA can be made from 75.0g of ammonium
formate and 125g of acetophenone? What mass of the
excess reactant remains?
Equation
Before
Change
After
CH5NO2
1.19 mol
+
C8H8O 
1.04 mol
C8H11N
0
+
CO2
0
+
H2O
0
# 6. How much PEA can be made from 75.0g of ammonium
formate and 125g of acetophenone? What mass of the
excess reactant remains?
Equation
CH5NO2
+
C8H8O 
C8H11N
+
CO2
+
H2O
Before
1.19 mol
1.04 mol
0
0
0
Change
-1.04 mol
-1.04 mol
+1.04 mol
+1.04 mol
+1.04 mol
After
0.15 mol
0 mol
1.04 mol
1.04 mol
1.04 mol
Looking at these numbers they all have coefficients of 1, so it’s easy to see
that the acetophenone is the one that runs out.
# 6. How much PEA can be made from 75.0g of ammonium
formate and 125g of acetophenone? What mass of the
excess reactant remains?
63.07 g AF
0.15 mol AF
x
= 9.46 g AF in excess
1 mol AF
121.20 g PEA
1.04 mol PEA
x
= 126 g PEA formed
1 mol PEA