Chemistry 20 – Lesson 21
Solution Stoichiometry
/80
1.
/6
Cu(s)
2 AgNO3 (aq) 
 Cu(NO3)2 (aq)
+
mCu  ?
+ 2 Ag (s)
cAgNO3  0.100 mol L
v AgNO3  0.250L
A. calculate moles
B. mole ratio
n AgNO3  0.100 mol L (0.250L)
C. calculate mass
n AgNO3
mCu  0.0125mol(63.55 g mol)
n AgNO3  0.0250mol
n Cu
2
1
0.0250mol n Cu

2
1
n Cu  0.0125mol

mCu  0.794 g
OR
mCu 
0.100 mol AgNO3
1L
 0.250 L 
1 mol Cu

2 mol AgNO3
63.55g Cu
1 mol Cu
mCu  0.794 g
2.
/7
2 Al(s)
m Al reacted
3 H2SO4 (aq) 
 3 H2 (g)
 m Alinitial  mAlafter cH2SO4  ?
+
+
Al2(SO4)3 (aq)
m Al  15.14g  9.74g
m Al  5.40g
A. Calculate moles
5.40g
26.98 g mol
n Al  0.200 mol
n Al 
B. Mole ratio
n H2SO4 n Al

3
2
n H2SO4 0.200 mol

3
2
n H2SO4  0.300 mol
C. Calculate concentration
0.300 mol
cH2SO4 
0.500 L
cH2SO4  0.600 mol L
OR
cH2SO4  3.40 g Al 
1mol Al 3mol H 2SO4
1


26.98g Al
2 mol Al
0.500 L
cH2SO4  0.600 mol L
Dr. Ron Licht
21 - 1
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3.
/8
Mg(s)
+
First calculate the mass of
2 HCl (aq)
cHCl  6.00 mol L
magnesium that reacted.
v HCl  2.40L


H2 (g)
+
MgCl2 (aq)
m Mg  ?
A. calculate moles
n HCl  6.00 mol L (2.40L)
B. mole ratio
C. calculate mass
n Mg
m Mg  7.20 mol(24.31 g mol)
n
 HCl
1
2
n Mg 14.4 mol

1
2
n Mg  7.20 mol
n HCl  14.4mol
m Mg reacted  175g
OR
mMg 
6.00 mol HCl
1mol Mg 24.32g Mg
 2.40L 

 175g
1L
2 mol HCl 1mol Mg
m Mg remaining  m Mg initial  m Mg reacted
m Mg remaining  200 g  175g
m Mg remaining  25g
4.
/5
2 NaBr (aq)
c NaBr  0.300
+
mol
L
v NaBr  0.120L
Cl2 (g) 

+
Br2 (l)
n Cl2  ?
A. calculate moles
n NaBr  0.300 mol L (0.120L)
n NaBr  0.0360mol
2 NaCl (aq)
B. mole ratio
n Cl2 n NaBr

1
2
n Cl2 0.0360mol

1
2
n Cl2  0.0180mol
OR
n Cl2 
1mol Cl2
0.300 mol NaBr
 0.120L 
1L
2 mol NaBr
n Cl2  0.0180 mol
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21 - 2
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5.
3 NH4OH (aq)
+
/8
c NH4OH  0.670 mol L


FeCl3 (aq)
3 NH4Cl (aq)
+
Fe(OH)3 (s)
mCu  ?
v NH4OH  0.050L
A. calculate moles
B. mole ratio
n NH4OH  0.670 mol L (0.050L)
C. calculate mass
n Fe(OH)3
m Fe(OH)3  0.0112mol(106.88 g mol)
n NH4OH  0.0335mol
1
n Fe(OH)3
1
n Fe(OH)3

n NH4OH
3
0.0335mol

3
 0.0112mol
m Fe(OH)3  1.193g
D. calculate % error
m exp yield  m theo yield
% error 
100%
m theo yield
% error 
1.135g  1.193g
100%
1.193g
% error  4.90%
6.
Mg (s)
+
2 HBr (aq)


MgBr2 (aq)
H2 (g)
m H2  ?
cHBr  1.35 mol L
/8
+
v HBr  0.0461L
A. calculate moles
n HBr  1.35 mol L (0.0461L)
n HBr  0.0622mol
B. mole ratio
n H2 n HBr

1
3
n H2 0.0622mol

1
2
n Fe(OH)3  0.0311mol
C. calculate mass
m H2  0.0311mol(2.02 g mol)
m H2  0.0629 g
D. calculate % error
m exp yield  m theo yield
% error 
100%
m theo yield
% error 
0.0556g  0.0629g
100%
0.0629g
% error  11.5%
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21 - 3
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7.
Na2SO4 (aq)
+
/8
c Na 2SO4  1.75 mol L
Ba(OH)2 (aq) 

2 NaOH (aq)
+
BaSO4(s)
m H2  ?
v Na 2SO4  0.0200L
A. calculate moles
n Na 2SO4  1.75 mol L (0.0200L)
n Na 2SO4  0.0350mol
B. mole ratio
n BaSO4 n Na 2SO4

1
1
n BaSO4 0.0350mol

1
1
n BaSO4  0.0350mol
C. calculate mass
m BaSO4  0.0350mol(233.40 g mol)
m BaSO4  8.17 g
D. calculate % yield
m exp yield
% yield 
100%
m theo yield
% yield 
6.17 g
100%
8.17 g
% yield  75.5%
8.
HNO3 (aq)
/6
cHNO3  3.00
+
mol
L
v HNO3  ?
c NaOH  0.10
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mol
NaNO3 (aq)
+
HOH (l)
m H2  ?
L
v NaOH  0.0600L
A. calculate moles
n NaOH  0.10 mol L (0.0600L)
n NaOH  0.00600mol


NaOH (aq)
B. mole ratio
n HNO3 n NaOH

1
1
n HNO3 0.00600mol

1
1
n HNO3  0.00600mol
21 - 4
C. calculate volume
0.00600mol
v HNO3 
3.00 mol L
v HNO3  2.00 mL
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9.
3 Ba(OH)2 (aq)
+
/8
cBa (OH)2  0.0450 mol L
Al2(SO4)3 (aq) 

3 BaSO4 (s)
+
2 Al(OH)3 (s)
mAl(OH)3  ?
mBaSO4  ?
v Ba (OH)2  0.0250L
A. calculate moles
B. mole ratios
n Ba (OH)2  0.0450 mol L (0.0250L) for BaSO
4
n Ba (OH)2  0.001125mol
n BaSO4 n Ba (OH)2

3
3
n BaSO4 0.001125mol

3
3
n BaSO4  0.001125mol
C. calculate masses
m BaSO4  0.001125mol(233.40 g mol)
m BaSO4  0.2626 g
m Al(OH)3  0.00075mol(78.01 g mol)
m Al(OH)3  0.0585g
for Al(OH)3
n Al(OH)3
2
n Al(OH)3
2
n Al(OH)3

n Ba (OH)2
3
0.001125mol

3
 0.000750mol
D. calculate % error
m exp yield  m theo yield
% error 
100%
m theo yield
% error 
0.31g  (0.2626g  0.0585g)
100%
(0.2626g  0.0585g)
% error  3.46%
10.
Pb(NO3)2 (aq)
+
2 RbBr (aq)


2 RbNO3 (aq)
PbBr2(s)
m PbBr2  m product filter paper  mfilter paper
mRbBr  ?
/6
+
m PbBr2  6.83g  0.21g
m PbBr2  6.62 g
A. calculate moles
6.62g
378.13 g mol
 0.0175mol
n PbBr2 
n PbBr2
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B. mole ratio
C. calculate mass
n RbBr n PbBr2

2
1
n RbBr 0.0175mol

2
1
n RbBr  0.0350mol
m RbBr  0.0350mol(165.37 g mol)
21 - 5
m RbBr  5.79 g
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11.
Ba(OH)2 (aq)
+
Mg(NO3)2 (aq) 

Ba(NO3)2 (aq)
+
mMg(OH)2  ?
mMg(NO3 )2  17.86g
/10
A. calculate moles
17.86 g
148.33 g mol
 0.1204mol
n Mg( NO3 )2 
n Mg( NO3 )2
B. mole ratio
n Mg(OH)2
1
n Mg(OH)2
1
n Mg(OH)2
Ba(OH)2 (aq)
+
Na2SO4 (aq)

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m Mg(OH)2  0.1204mol(58.33 g mol)
1
0.1204 mol

1
 0.1204 mol
n Na2SO4  0.035mol
A. mole ratio
n BaSO4 n Na 2SO4

1
1
n BaSO4 0.035 mol

1
1
n BaSO4  0.035 mol
C. calculate mass
n Mg( NO3 )2


Mg(OH)2 (s)
BaSO4 (s)
m Mg(OH)2  7.023g
+
2 NaOH (aq)
mBaSO4  ?
B. calculate mass
m BaSO4  0.035 mol(233.40 g mol)
m BaSO4  8.2 g
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