Notes
Chapter 9 – Stoichiometry
Interpret the following equation:
N2 (g) +
Notes
3H2 (g) →
2NH3 (g)
in terms of:
makes
Numbers of moles
→
1 mol N2
+
3 mol H2
Numbers of molecules→
6.02 X 1023 molecules N2 + 3(6.02 X 1023) molecules H2 →
→
yields
2 moles NH3
2(6.02 X 1023) molecules NH3
Volumes of gases
→
Or
22.4 L of N2
22.4 L of N2
+
+
3(22.4 L) of H2
67.2 L of H2
→
→
2(22.4L) NH3
44.8 L of NH3
Masses of gases
→
Or
28 g of N2
28 g of N2
+
+
3 ( 2g) of H2 →
6 g of H2
→
2 ( 17 g) of NH3
34 g of NH3
We said above that 1 mole of N2 will produce 2 moles of NH3. But what if I give you 0.6 moles
of N2? How many moles of NH3 will you make?
N2 (g) +
0.6 mol
3H2 (g) →
Use dimensional analysis and
begin with the number you were
given in the problem.
2NH3 (g)
? mol
0.6 mol N2
Work from moles of N2 to
moles of NH3.
2 mol NH3
1 mol N2
=
1.2 mol NH3
In the equation below, how many moles of Aluminum are needed to form 3.7 moles of Al2O3?
4Al
+
? mol
3O2
Use dimensional analysis and
begin with the number you were
given in the problem.
→
2Al2O3
3.7 mol
3.7 mol Al2O3
Work from moles of Al2O3 to
moles of Al.
4 mol Al
2 mol Al2O3
= 7.4 mol Al
In the above equation, how many moles of Oxygen are required to react completely with 14.8
moles of Aluminum?
4Al
+
14.8 mol
3O2
→
? mol
Use dimensional analysis and
begin with the number you were
given in the problem.
2Al2O3
14.8 mol Al
Work from moles of Al to
moles of O2.
3 mol O2
4 mol Al
= 11.1 mol O2
How many moles of Al2O3 are formed when 0.78 mol O2 reacts with Aluminum?
4Al
+
3O2
→
0.78 mol
Use dimensional analysis and
begin with the number you were
given in the problem.
0.78 mol O2
2Al2O3
? mol
Work from moles of O2 to
moles of Al2O3.
2 mol Al2O3
3 mol O2
= 0.52 mol Al2O3
Notes
Problem:
Chapter 9 – Stoichiometry
Notes
Calculate the number of grams of NH3 produced by the reaction of 5.40 g of
hydrogen with an excess of nitrogen.
Note: you cannot move from grams to grams, only from moles to moles.
So, you must work from grams H2 →
N2 (g) +
moles H2 →
3H2 (g)
5.4 g
→
? grams
(4)
moles
moles
(2)
(1)
2nd, map out the path across
the top of your dimensional
analysis grid.
grams NH3.
2NH3 (g)
(1)
1st, draw your path with arrows.
moles NH3 →
5.4 g H2
(3)
(2)
(3)
1 mol H2
2 g H2
(4)
2 mol NH3
3 mol H2
17 g NH3 =
1 mol NH3
= 30.6 g NH3
Note: between moles and grams, mole is always 1,
grams is the total mass from the periodic chart.
Note: between moles and moles, use the coefficients.
Problem:
Acetylene gas (C2H2) is produced by adding water to Calcium carbide (CaC2).
How many grams of Acetylene are produced by adding water to 5.00 g of CaC2?
1st, draw your path with arrows.
CaC2 (s)+
5.00 g
2H2O (l)
→
C2H2 (g)+
?g
(1)
Ca(OH)2 (aq)
(4)
moles
moles
(2)
(3)
(1)
2nd, map out the path across
the top of your dimensional
analysis grid.
3rd, do the math.
(2)
5.00 g CaC2 1 mol CaC2
64 g CaC2
(3)
1 mol C2H2
3 mol CaC2
(4)
26 g C2H2 =
1 mol C2H2
= 2.03 g C2H2
Notes
Problem:
Chapter 9 – Stoichiometry
Notes
How many grams of CaC2 are needed to react completely with 49.0 g of H2O?
1st, draw your path with arrows.
CaC2 (s)+
?g
2H2O (l)
C2H2 (g)+
→
49.00 g
(4)
Ca(OH)2 (aq)
(1)
moles
(3)
moles
(2)
(1)
2nd, map out the path across
the top of your dimensional
analysis grid.
(2)
49.00 g H2O 1 mol H2O
18 g H2O
(3)
1 mol CaC2
2 mol H2O
3rd, do the math.
Problem:
(4)
64 g CaC2 =
1 mol CaC2
= 87.11 g CaC2
Rust (Fe2O3) is produced when Iron (Fe) reacts with Oxygen (O2). How many
grams of Fe2O3 are produced when 12.0 g of Iron rusts?
1st, draw your path with arrows.
4Fe
+
3O2
→
12.00 g
2Fe2O3
?g
(1)
(4)
moles
moles
(2)
(3)
(1)
2nd, map out the path across
the top of your dimensional
analysis grid.
3rd, do the math.
12.00 g Fe
(2)
1 mol Fe
56 g Fe
(3)
2 mol Fe2O3
4 mol Fe
(4)
160 g Fe2O3 =
1 mol Fe2O3
= 17.14 g Fe2O3