Chapter(s) 12 Name _____________________
Period _____
Stoy
Set I. Conversion Factors
– kee – ahm – eh - tree
Consider the following reaction:
1. 3 Hg(OH) a) c)
2
+ 2 H
2 mole H
3
PO
3
4
PO
4
3 moles Hg(OH)
 Hg b) 3 moles Hg(OH)
2
2
3
(PO
4
= 6 mole H
2
)
2
+ 6 H
2
O
O
= 1 mole Hg
= 2 moles H
3
(PO
3
PO
4
4
)
2
What is the mole to mole relationship for phosphoric acid to water?
What is the mole to mole relationship for Hg(OH)
Hg
3
(PO
4
)
2
?
2
to
What is the mole to mole relationship for Hg(OH)
2
to d) 1 mole Hg
3
(PO
4
)
2
= 6 moles H
2
O
H
H
3
2
PO
4
O?
?
What is the mole to mole relationship for Hg
3
(PO
4
)
2
to
2.
_2_ SO
2
+ __O
2
 _2_ SO
3
2. 2 mol SO
1 mole O
2
= 1 mole O
2
, 2 moles SO
2
= 2 moles SO
3,
2
= 2 mole SO
3
3.
__PCl
3
+ __Cl
2
 __PCl
5
3. Everything is 1:1
4.
_4_ NH
3
+ _3_ O
2
 _2_ N
2
+ _6_ H
2
O
4. 4 mol NH
3
= 3 mol O
2
, 4 mol NH
3
= 2 mol N
2
,
4 mol NH
3
= 6 mol H
2
3 mol O
2
= 6 mol H
2
0, 3 mol O
O, 2 mol N
2
2
= 2 mol N
2
,
= 6 mol H
2
O
5.
__Fe
2
O
3
+ _3_ CO  _2_ Fe + _3_ CO
2
5. 1 mol Fe
2
O
3
= 3 mol CO, 1 mol Fe
2
O
3
= 2 mol Fe,
1 mol Fe
2
O
3
=3 mol CO
3 mol CO = 3 mol CO
2,
3 mol CO = 2 mol Fe,
2
, 2 mol Fe = 3 mol CO
2
Set II. Mole to mole and mole to mass
On a separate sheet of paper balance the equations for each problem and solve the questions. Don’t forget to make sure the reaction is balanced!
S + O
2
 SO
2
1.
“sulfur burns in oxygen gas to form sulfur dioxide” a.
1.56 mol S b.
10 mol S c.
1.93 x 10 25 mc O
2
How many moles of sulfur must be burned to give 100.0 g of SO
2
?
How many moles of sulfur must be burned to completely react with 10 moles of oxygen?
If there are 32 moles of oxygen, how many molecules of oxygen are there?
2.
BaO + 2 HNO
3
 Ba(NO
3
)
2
+ H
2 a.
b.
2.56 x 10
14.9
7 mg Ba(NO mol HNO
3
3.
Cu + 3AgNO
3
 Cu(NO a.
9.85 mol AgNO
3 b.
15.5 mole Cu
3
)
2
3
)
3
O
How many mg of Ba(NO
+ 3Ag
3
)
2
can be formed from 196. moles of HNO
How many moles of AgNO
3
are needed to produce 9.85 mol of silver.
How many moles of copper produce 5.00 kg of silver?
3
?
If 7.45 moles of water are formed during this reaction, how many moles of HNO were used?
3

4.
2 NaCl + Zn(NO
3
) a.
3 mol ZnCl
2
2 b.
0.565 mol ZnCl
Set II Extra Practice

2
ZnCl
2
+ 2 NaNO
3
6 moles NaCl should produce how many moles of ZnCl
How many moles of ZnCl
2
2
?
can be produced from 107 g of Zn(NO
3
)
2
?
6 NaOH + 2 Al  2 Na
3
AlO
3
+ 3 H
2 a.
How many moles of hydrogen will be produced if 2 moles of NaOH react will excess Al?
1 mol H
2 b.
How many moles of hydrogen can be prepared from 1.0 gram of aluminum?
0.056 mol H
2 c.
If 3 moles of Na
3
AlO
3 are produced, how many moles of Al were present before the reaction if the aluminum all reacted?
3 mol Al d.
How many kg of Na
3
7.92 kg Na
3
AlO
3
AlO
3
can be formed from 165. moles of sodium hydroxide? e.
How many moles of NaOH are required to produce 3.00 g of hydrogen?
2.98 mol NaOH
Set II. Mole to mole and mole to mass Worked Out
On a separate sheet of paper balance the equations for each problem and solve the questions. Don’t forget to make sure the reaction is balanced!
1.
“sulfur burns in oxygen gas to form sulfur dioxide”
S + O
2
 SO
2 b.
How many moles of sulfur must be burned to give 100.0 g of SO
2
?
G: 100.0 g SO
2
W: moles S
R: 1 mol S = 1 mol SO
1 mol SO
2
2
= 64.064 g SO
2
100 g SO
2 x
1 mol SO
2 x
1 64.064 g SO
2
1 mol S
1 mol SO
=
1.56 mol S
2 c.
How many moles of sulfur must be burned to completely react with 10 moles of oxygen?
G: 10 mol O
W: mol S
2
R: 1 mol S = 1 mol O
2
10 mol O
2 x
1 mol S
1 1 mol O
2
=
10 mol S d.
If there are 32 moles of oxygen, how many molecules of oxygen are there?
G: 32 mol O
W: mc O
2
2
R: 1 mol = 6.02 x 10
32 mol O
2 x
23 mc
6.02 x 10
1 1 mol
23 mc
=
1.93 x 10 25 mc O
2
2. BaO + 2 HNO
3
 Ba(NO
3
)
2
+ H
2
O

a.
How many mg of Ba(NO
3
)
G: 196.0 moles HNO
3
W: mg of Ba(NO
3
)
2
R: 1g = 1000 mg
1 mole Ba(NO
3
)
2
can be formed from 196.0 moles of HNO
1 mole Ba(NO
3
)
2
2
= 261.338 g Ba(NO
3
)
2
= 2 moles HNO
3
196.0 moles HNO
3 x
1 mole Ba(NO
3
?
3
)
2 x
261.338 g Ba(NO
3
)
2 x
1000 mg
=
2.56 x 10 7 mg Ba(NO
3
)
2
1 2 moles HNO
3
1 mole Ba(NO
3
)
2
1g b.
If 7.45 moles of water are formed during this reaction, how many moles of HNO
G: 7.45 mol water
W: moles HNO
R: 1 mole H
7.45 mol H
2
2
3
O = 2 HNO
3
3
were used?
O x
2 mol HNO
1 1 mole H
2.
Cu + 3AgNO
3
 Cu(NO
2
O`
3
)
3 c.
How many moles of AgNO
3 =
14.9
mol HNO
+ 3Ag
3
3
are needed to produce 9.85 mol of silver.
G: 9.85 mol of silver
W: moles of AgNO
3
R: 3 mol Ag = 3 mol AgNO
3
9.85 mol Ag x 3 mol AgNO
3 =
9.85 mol AgNO
3
1 3 mol Ag d.
How many moles of copper produce 5 kg of silver?
G:
5 kg of silver
W: moles of copper
R: 1 kg = 1000 g
1 mole Cu = 3 mole Ag
1 mole Ag = 107.868 g Ag
5 kg Ag x
1000 g
x
1 mole Ag
)
2
 ZnCl
2
+ 2 NaNO
3 a.
6 moles NaCl should produce how many moles of ZnCl
2
?
G: 6 moles NaCl
W: moles of ZnCl
2
R: 2 moles NaCl = 1 mole ZnCl
2 x
1 mole Cu
=
15.5 mole Cu
1 1 kg 107.868 g Ag 3 mole Ag
5. 2 NaCl + Zn(NO
3
6 moles NaCl x
1 mole ZnCl
2 =
3 mol ZnCl
2
1 2 moles NaCl b.
How many moles of ZnCl
2
can be produced from 107.0 g of Zn(NO
3
)
2
?
G: 107.0 g of Zn(NO
3
)
2
W: moles of ZnCl
2
R: 1 mole Zn(NO
3
= 189.398 g Zn(NO
3
)
)
2
= 1 mole ZnCl
2
1 mole Zn(NO
3
)
2
107.0 g Zn(NO
3
)
2 x
1 mole Zn(NO
3
)
1 189.398 g Zn(NO
3
Extra Practice
2
2 x
)
2
1 mole ZnCl
2 =
0.565 mol ZnCl
1 mole Zn(NO
3
)
2
2

6 NaOH + 2 Al  2 Na
3
AlO
3
+ 3 H
2 a.
How many moles of hydrogen will be produced if 2 moles of NaOH react will excess Al?
G: 2 mol NaOH
W: moles of H
2
R: 6 moles NaOH = 3 moles H
2
2 mol NaOH x
3 mole H
2
= 1.00 mol H
1 6 mol NaOH
2 b.
How many moles of hydrogen can be prepared from 1 gram of aluminum?
G: 1 gram Al
W: moles ofH
2
R: 2 mol Al = 3 mol H
2
1 mole Al = 189.398 g Zn(NO
3
)
2
1 g Al x 1 mol Al x 3 mol H
2
1 26.982 g Al 2 mol Al
= 0.0556 mol H
2 c.
If 3 moles of Na
G: 3 mol Na
3
AlO
3
3
AlO
3 are produced, how many moles of Al were present before the reaction if the aluminum all reacted?
W: moles Al
R: 2 mol Na
3
AlO
3
= 2 mol Al
3 mol Na
3
AlO
3 x
2 mol Al
1 2 mol Na
3
AlO
3
= 3 mol Al d.
How many kg of Na
3
AlO
3
can be formed from 165.0 moles of sodium hydroxide?
G: 165.0 mol NaOH
W: kg Na
3
AlO
3
R: 6 mol NaOH = 2 mol Na
1 mole Na
3
AlO
3
3
AlO
3
= 120.959 g Na
3
AlO
3
165 mol NaOH x 2 mol Na
3
AlO
3
x 143.949 g Na
3
AlO
1 6 mol NaOH 1 mol Na
3
AlO
3
3
x 1 kg = 7.92 kg Na
1000 g
3
AlO
3 e.
How many moles of NaOH are required to produce 3 g of hydrogen?
G: 3 g H
2
W: mol NaOH
R: 6 mol NaOH = 3 mol H
2
1 mol H
2
= 2.016 g H
2
3 g H
2
x 1 mol H
1 2.016 g H
2
x 6 mol NaOH = 2.98 mol NaOH
2
3 mol H
2
Set III. Mass to mass and more
On a separate sheet of paper balance the equations for each problem and solve the questions.

1.
NaCl + AgNO
3
 AgCl + NaNO a.
191 g AgCl b.
227g AgNO
3 c.
25.0g NaNO
3 d.
4 mol NaNO e.
287 g AgCl
3 f.
0.35 mol AgNO
3
3
78.0 g of NaCl should produce how many grams of AgCl?
78.0 g of NaCl are required to react completely with how many grams of silver nitrate?
If 50.0 grams of silver nitrate are consumed in a reaction, how many grams of sodium nitrate are produced with it?
How many moles of sodium nitrate will form from 4 moles of sodium chloride?
How many grams of silver chloride will form from 2.00 moles of sodium chloride?
If there are 60. g of silver nitrate how many moles of silver nitrate are there?
Set III Extra Practice
6 NaOH + 2 Al  2 Na
3
AlO
3
+ 3 H
2 a.
5.79 mol Al b.
0.028 kg NaOH c.
d.
e.
1.9 mol NaOH
10.5 mol H
2
How many moles of aluminum are required to produce 17.5 g of hydrogen?
How many kg of NaOH are required to react with 6.5 grams of Al?
29.9 g NaOH & 6.72g Al If 42.3 g of Na
3
AlO
3
are produced in the reaction, how many g. of NaOH and Al are used?
How many moles of sodium hydroxide are in 76 g of sodium hydroxide?
How many moles of hydrogen are released if 7.00 moles of aluminum are consumed?
Set III. Mass to mass and more Worked Out
On a separate sheet of paper balance the equations for each problem and solve the questions.
1.
NaCl + AgNO
3
 AgCl + NaNO
3 a.
78.00 g of NaCl should produce how many grams of AgCl?
G: 78.00 g NaCl
W: g AgCl
R: 1 mol NaCl = 1 mol AgCl
1 mol NaCl = 58.443g NaCl
1 mol AgCl = 143.321 g AgCl
78.00g NaCl x 1 mol NaCl x 1 mol AgCl x 143.321 g AgCl = 191.28g AgCl
1 58.443 g NaCl 1 mol NaCl 1 mol AgCl b.
78.00 g of NaCl are required to react completely with how many grams of silver nitrate?
G: 78.00 g NaCl
W: g AgNO
3
R: 1 mol NaCl = 1 mol AgNO
3
1 mol NaCl = 58.442 g NaCl
1 mol AgNO
3
= 169.872 g AgNO
3
78.00g NaCl x 1 mol NaCl x 1 mol AgNO
3
x 169.872g AgNO
3
= 227g AgNO
3
1 58.443 g NaCl 1 mol NaCl 1 mol AgNO
3 c.
If 50.0 grams of silver nitrate are consumed in a reaction, how many grams of sodium nitrate are produced with it?
G: 50.0 g AgNO
3
W: g NaNO
3
R: 1 mol AgNO3 = 1 mol NaNO
1 mol AgNO
3
= 169.872 g AgNO
1 mol NaNO
3
= 84.994 g AgNO
3
50.0g AgNO
3
x 1 mol AgNO
3
3
3
x 1 mol NaNO
3
x 84.994g NaNO
3
= 25.01g NaNO
3

1 169.872g AgNO
3
1 mol AgNO
3
1 mol NaNO
3 d.
How many moles of sodium nitrate will form from 4 moles of sodium chloride?
G: 4 mol NaCl
W: mol AgNO3
R: 1 mol NaCl = 1 mol AgNO
3
4 mol NaCl x 1 mol AgNO
3
= 4 mol AgNO
3
1 1 mol NaCl e.
How many grams of silver chloride will form from 2 moles of sodium chloride?
G: 2 mol NaCl
W: g AgCl
R: 1 mol AgCl = 1 mol NaCl
1 mol AgCl = 143.321 g AgCl
2 mol NaCl x 1 mol AgCl x 143.321g AgCl = 287g Ag
1 1 mol NaCl 1 mol AgCl f.
If there are 60 g of silver nitrate how many moles of silver nitrate are there?
G: 60g AgNO
3
W: 1 mol AgNO
60.0g AgNO
3
3
= 169.872 g AgNO
3
x 1 mol AgNO
3
= 0.350 mol AgNO
3
1 169.872 g AgNO
3
Set III Extra Practice
6 NaOH + 2 Al  2 Na
G: 17.5 g H
2
3
AlO
3
+ 3 H
2 a.
How many moles of aluminum are required to produce 17.5 g of hydrogen?
W: mol Al
R: 3 mol H
1 mol H
17.5 g H
2
2
2
= 2 mol Al
= 2.016 g H
2
x 1 mol H
2
x 2 mol Al = 5.787 mol Al
1 2.016 g H
2
3 mol H
2 b.
How many kg of NaOH are required to react with 6.5 grams of Al?
G: 6.5 g Al
W: kg NaOH
R: 2 mol Al = 6 mol NaOH
1 mol NaOH = 39.997 g NaOH
6.5 g Al x 1 mol Al x 6 mol NaOH x 39.997 g NaOH x 1 kg NaOH = 0.0289 kg NaOH
1 26.982 g Al 2 mol Al 1 mol NaOH 1000 g NaOH c.
If 42.3 g of Na
3
AlO
3
are produced in the reaction, how many g. of NaOH and Al are used?
G: 42.3 g AgNO
W: g NaOH
3
R: 6 mol NaOH = 2 mol AgNO
3

1 mol AgNO
3
= 169.872 g AgNO
3
1 mol NaOH = 39.997 g NaOH
42.3 g AgNO
3
x 1 mol AgNO
3
x 6 mol NaOH x 39.997 g NaOH = 29.9 g NaOH
1 169.872 g AgNO
3
2 mol AgNO
3
1 mol NaOH
3
x 1 mol AgNO
3
x 2 mol Al x 26.982 g Al = 6.72 g Al
1 169.872 g AgNO
3
2 mol AgNO
3
1 mol Al d.
How many moles of sodium hydroxide are in 76 g of sodium hydroxide?
G: 76 g NaOH
W: mol NaOH
R: 1 mol NaOH = 39.997g
76 g NaOH x 1 mol NaOH = 1.90 mol NaOH
1 39.997g NaOH e.
How many moles of hydrogen are released if 7 moles of aluminum are consumed?
G: 7 mol Al
W: mol H
2
G: 42.3 g AgNO
W: g Al
3
1 mol AgNO
3
R: 6 mol NaOH = 2 mol AgNO
3
= 169.872 g AgNO
1 mol NaOH = 26.982 g Al
42.3 g AgNO
3
2
R: 2 mol Al = 3 mol H
7 mol Al x 3 mol H
2
= 10.5 mol H
1 2 mol Al
2
Set IV. Molecules and formula units to mass and more
On a separate sheet of paper balance the equations for each problem and solve the questions.
1.
Na
2
O + H
2
O  2 NaOH a. How many grams of NaOH are produced from 2.01 x 10 23
26.7g NaOH b. How many formula units of Na
1.2 x 10 22 fun Na
2
2.61 x 10 24 mc H
2
O
O molecules of water?
2
O are required to produce 1.6 grams of NaOH? c. How many molecules of water are needed to react completely with 4.35 moles of sodium oxide?
2.
CH
4
+ 2 O a. If 30.0 g of methane are combusted completely, how many milligrams of water are produced?
6.74 x 10 4
2
 CO
2
+ 2 H
2
O (when heated)
2 are needed to completely combust with 4.00 moles of methane?
255 g O
2 mg H
2
O b. How many grams of O c. How many molecules of Carbon dioxide are formed from 2.90 x 10
2.90 x 10 22 mc CO
Extra Practice
2
22 molecules of methane?

3.
Cu + 3 AgNO
24
3
 Cu(NO
atoms Ag
3
)
3
+ 3 Ag a.
How many atoms of Ag are produced when 2.44 moles of Cu are consumed?
4.41 x 10 b.
How many atoms of copper produce 4.22 x 10 22
1.41 x 10 22 atoms Cu
formula units of silver?
4. BaO + H
2
SO
4
 BaSO
fun BaSO
4
+ H
2
O a. How many formula units of BaSO
4 are present if there are 43 g of BaSO
4
?
1.1 x 10 23
4 b. How many moles of water are formed from 6.04 x 10 20 molecules of sulfuric acid?
1.00 x 10 -3 mol H
2
O c. How many formula units of BaSO
4
23 fun BaO
can be produced from 3.01 x 10 23 formula units of BaO?
3.01 x 10 d. How many formula units of barium oxide are used if 26.0 g of water are produced?
8.69 x 10 23 fun BaO e. If 4 moles of water are formed how many formula units of Barium oxide were consumed?
2.41 x 10 24 fun BaO
Set IV. Molecules and formula units to mass and more Worked Out
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. Na
2
O + H a. How many grams of NaOH are produced from 2.01 x 10
2.01 x 10
2
O  2 NaOH
23 mc H
2
O x 1 mol H
1 6.02 x 10 23
2 b. How many formula units of Na
O x 2 mol NaOH x 39.997g NaOH = 26.7g NaOH
mc H
2
O 1 mol H
2
23 molecules of water?
O 1 mol NaOH
2
O are required to produce 1.60 grams of NaOH?
1.60g NaOH x 1 mol NaOH x 1 mol Na
2
O x 6.02 x 10 23 fun Na
O
2
O = 1.20 x 10 22 fun Na
2
O
1 39.997 NaOH 2 mol NaOH 1 mol Na
2 e.
How many molecules of water are needed to react completely with 4.35 moles of sodium oxide?
mc H
2
O 4.35 mol Na
2
O x 1 mol H
2
1 1 mol Na
2
O x 6.02 x 10 23 fun Na
2
O = 2.61 x 10 24
O 1 mol Na
2
O
2. CH
4
+ 2 O
2
 CO
2
+ 2 H
30.0 g CH
2
O (when heated) a. If 30.0 g of methane are combusted completely, how many milligrams of water are produced?
4
x 1 mol CH
4
x 2 mol H
2
O x 18.015 g H
2
O x 1000 mg = 67375= 6.74 x 10 4 mg H
2
O
1 mol H
2
O 1 g 1 16.043 g CH
4
1 mol CH
4 b. How many grams of O
2 are needed to completely combust with 4 moles of methane?
4 mol CH
4
x 2 mol O
2
x 31.998 g O
2
1 1 mol CH
4
1 mol O
2
= 255 g O
2 f.
How many molecules of Carbon dioxide are formed from 2.90 x 10 22
2.90 x 10 22 mc CH
4
x 1 mol CH
4
x 1 mol CO
2
x 6.02 x 10 23 molecules of methane?
mc CO
2
= 2.90 x 10 22 mc CO
2

1 6.02 x 10 23
3. Cu + 3 AgNO
3
 Cu(NO
3
)
3
mc CH
+ 3 Ag
4
1 mol CH
4
1 mole CO
2 c.
How many atoms of Ag are produced when 2.44 moles of Cu are consumed?
2.44 mol Cu x 3 mol Ag x 6.02 x 10 23 atoms Ag = 4.41 x 10
1 1 mol Cu 1 mol Ag
24 atoms Ag d.
How many atoms of copper produce 4.22 x 10 22 formula units of silver?
4.22 x 10 22 fun Ag x 1 mol Ag x 1 mol Cu x 6.02 x 10 23
1 6.02 x 10 23 atoms Cu = 1.41 x 10 22 atoms Cu
fun Ag 3 mol Ag 1 mol Cu
Extra Practice
BaO + H
2
SO
4
 BaSO
4
+ H
2
O a. How many formula units of BaSO
43 g BaSO
4
x 1 mol BaSO
4
4 are present if there are 43 g of BaSO
x 6.02 x 10 23 = 1.11 x 10
4
23
?
fun BaSO
4 fun BaSO
4
1 233.392 g BaSO
4
1 mol BaSO
4 b. How many moles of water are formed from 6.04 x 10
6.04 x 10 20 mc H
2
SO
4
x 1 mol H
2
SO
1 6.02 x 10 23 mc H
4
2
SO
20 molecules of sulfuric acid?
x 1 mol H
4
1 mol H
2
O = 1.00 x 10 -3 mol H
2
SO
4
2
O c. How many formula units of BaSO
3.01 x 10
1 6.02 x 10
26g H
2
23 fun BaO x 1 mol BaO x 1 mol BaSO
O x 1 mol H
2
1 18.015g H
2
23
4
can be produced from 3.01 x 10
4
x 6.02 x 10 23
23 formula units of BaO?
fun BaSO
fun BaO 1 mol BaO 1 mol BaSO
4
4
= 3.01 x 10 d. How many formula units of barium oxide are used if 26 g of water are produced?
23
O x 1 mol BaO x 6.02 x 10
O 1 mol H
23 fun BaO = 8.69 x 10
2
O 1 mol BaO
23 fun BaO g.
If 4 moles of water are formed how many formula units of Barium oxide were consumed?
fun BaO = 2.41 x 10 24 fun BaO 4 mol H
2
O x 1 mol BaO x 6.02 x 10 23
1 1 mol H
2
O 1 mol BaO
fun BaO
Set V. Molar Volume Conversions
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. N
2
(g) + 3 H
2
(g)  2 NH a.
1.49 mol N
2
3
(g)
How many mol of N
2 gas are needed to react with 100.0 L of H
2 gas to produce NH
3
gas? b.
59.7 L NH c.
36.4 L N d.
6 L NH g.
3 f.
10.5 L H
2 e.
3.61 x 10
2
1.16 x 10
3
24 mc N
23
How many liters of ammonia are produced when 89.6 L of H
If there are 9.78 x 10
If 9 liters of H
2
mc N
2
2
23
2
are used? mc of nitrogen gas, how many liters of N
2
are there? are used, how many liters of ammonia are produced?
If 12 moles of NH
3 are produced, how many molecules of nitrogen were used?
If 7.00 liters of ammonia are produced, how many liters of H
2
were consumed?
If 13 liters of hydrogen were used, how many molecules of nitrogen reacted?
Set V. Molar Volume Conversions Worked Out
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. N
2
(g) + H
2
(g)  NH
3 a.
How many mol of N
(g)
2 gas are needed to react with 100.0 L of H
G: 100L H
2
W: mol N
2
2 gas to produce NH
3
gas?

R: 1 mol H
100.0 L H
2
2
= 22.4 L H
x 1 mol H
2
2
x 1 mol N
2
1 22.4L H
2
3 mol H
2
= 1.49 mol N
2 b.
How many liters of ammonia are produced when 89.6 L of H
G: 89.6 L H
2
W: L NH
3
2
are used?
R: 1 mol H
2
= 22.4 L H
1 mol NH
3
89.6 L H
2
x 1 mol H
1 22.4L H
2
2
= 22.4 L NH
2 mol NH
3
= 3 mol H
2
3
2
x 2 mol NH
3
x 22.4 L NH
3 mol H
2
3
1 mol NH
3
= 59.7 L NH
3
OR 89.6 L H
2
x 2 L NH
3
= 59.7 L NH
3
1 3 L H
2 c.
If there are 9.78 x 10 23 mc of nitrogen gas, how many liters of N
2
are there?
G: 9.78 x 10 23 mc N
2
W: L N
2
R: 1 mol N
2
= 6.02 x 10 23 mc N
2
1 mol N
2
= 22.4 L N
2
9.78 x 10 23 mc N
2 x 1 mol N
2 x 22.4 L N
mc N
2
1 mol N
2
= 36.4 L N
2
2
1 6.02 x 10 23 d.
If 9 liters of H
2 are used, how many liters of ammonia are produced?
G: 9L H
2
W: L NH
3
R: 1 mol H
2
= 22.4 L H
1 mol NH
2
3
= 22.4L NH
3
OR 3L H
3 mol H
2
= 2 mol NH
3
9L H
2
x 1 mol H
2
x 2 mol NH
3
x 22.4L NH
1 mol NH
3
2
= 2 L NH
3
= 6L NH
3
3
1 22.4 L H
2
3 mol H
2
OR 9L H
2
x 2L NH
3
= 6L NH
3
1 3L H
2 e.
If 12 moles of NH
3 are produced, how many molecules of nitrogen were used?
G: 12 mol NH
3
W: mc N
2
R: 2 mol NH
3
= 1 mol N
1 mol N
2
2
= 6.02 x 1023 mc N
2
12 mol NH
3
x 1 mol N f.
If 7 liters of ammonia are produced, how many liters of H
2
1 2 mol NH
x 6.02 x 10
3
23 mc N
1 mol N
2
2
= 3.61 x 10
2
24 mc N
2
were consumed?
G: 7 L NH
3
W: L H
2
R: 2 mol NH
1 mol H
2
3
= 3 mol H
= 22.4L H
2
2
1 mol NH
3
= 22.4 L NH
3
OR 2 L NH
3
= 3 L H
2

1 2 L NH g.
If 13 liters of hydrogen were used, how many molecules of nitrogen reacted?
G: 13L H
W: mc N
2
2
R: 3 mol H
1 mol H
2
2
= 1 mol N
2
mc N
2
7L NH
3
x 1 mol NH
3
x 3 mol H
2
x 22.4L H
1 22.4 L NH
3
2 mol NH
3
OR 7L NH
3
x 3 L H
1 mol H
2
= 10.5L H
2
3
2
2
= 10.5L H
2
1 mol N
13 L H
2
2
= 22.4 L H
= 6.02 x 10
x 1 mol H
1 22.4L H
2
23
2
x 1 mol N
2
x 6.02 x 10
2
3 mol H
2
1 mol N
2
23 mc N
2
= 1.16 x 10 23 mc N
2
Set VI. Limiting Reactants and Excess Reactants
On a separate sheet of paper balance the equations for each problem and solve the questions.
1.
When 1.21 mol of solid zinc react with 2.10 mol hydrochloric acid, which reactant will be in excess and by how much? (write a balanced reaction first)
2.
Zinc and 0.160 mol are in excess
When 0.015 mol of Zinc metal reacts with 6.25 x 10 -3 mol lead (II) nitrate solution, crystals of lead form along with zinc nitrate. Which reactant is the limiting reactant?
Lead (II) nitrate
3.
If 0.135 mol of iron metal react with 0.100 mol hydrochloric acid solution, hydrogen gas and iron (II) chloride, are obtained. Which reactant will be in excess and by how much?
Fe = excess reactant; 0.085 moles are in excess
4.
If 0.387 mol of aluminum metal are reacted with 0.417 mol of copper (II) sulfate, then aluminum sulfate and copper are formed. Which reactant is the limiting reactant? Calculate the moles of the excess. (Reaction:2Al + 3CuSO
4
 3Cu +Al lim reac = copper (II) sulfate
amount of excess: 0.109 mol Al
2
(SO
4
)
3
)
Set VI. Limiting Reactants Worked Out
*Show your work for each problem & don’t forget to start off by writing a balanced reaction!*
1.
When 1.21 mol of solid zinc react with 2.10 mol hydrochloric acid, which reactant will be in excess and by how much? (write a balanced reaction first) Zinc and 0.16 mol are in excess
Zn + 2HCl  ZnCl
2
+ H
2
2.10 mol HCl 1 mol Zn = 1.05 mol Zn
2 mol HCl needed excess
1.21 mol Zn 2 mol HCl = 2.42 mol HCl available 1 mol Zn needed limiting
Moles of excess: 1.21 available – 1.05 mol required = 0.16 mol of excess
2.
When 0.015 mol of Zinc metal reacts with 6.25 x 10 -3 mol lead (II) nitrate solution, crystals of lead form along with zinc nitrate. Which reactant is the limiting reactant? Lead (II) nitrate

Zn + Pb(NO
3
)
2
 Pb + Zn(NO
3
)
2 available
0.015 mol Zn
0.00625 mol Pb(NO
3
)
2 needed
0.015 mol Pb(NO
3
0.00625 mol Zn
)
2 lead (II) nitrate = limiting zinc = excess
3.
If 0.135 mol of iron metal react with 0.100 mol hydrochloric acid solution, hydrogen gas and iron (II) chloride, are obtained. Which reactant is the limiting reactant? And, how many moles of hydrogen was is formed? Fe = excess; 0.085 mol in excess
Fe + 2HCl  H
2
+ FeCl
2 available
1 mol Fe needed
0.135 mol Fe 2 mol HCl = 0.271 mol HCl
Limiting reactant
0.100 mol HCl 1 mol Fe = 0.0500mol Fe
2 mol HCl Excess
0.135 – 0.0500= 0.085 mols in excess
4.
If 0.387 mol of aluminum metal are reacted with 0.417 mol of copper (II) sulfate, then aluminum sulfate and copper are formed. Which reactant is the limiting reactant? Calculate the moles of the excess. lim reac = copper (II) sulfate
2Al + 3CuSO
4 amount of excess: 0.109 mol Al
 3 Cu +Al
2
(SO
4
)
3
0.387 mol Al 3 mol CuSO
2 mol Al
4
= 0.581 mol CuSO
4 limiting
0.417 mol CuSO
4
2 mol Al = 0.278 mol Al excess
3 mol CuSO
4
0.387 mol Al – 0.278 mol Al = 0.109 mol Al excess
Set VII. Theoretical yield and percent yield.
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. Calculate the percent yield if the theoretical yield = 128 g, and actual yield = 112 g
87.5%
2. Calculate the actual yield if the theoretical yield = 23.98 kg, and the percent yield = 78.4%
18.8 kg = actual
3. I react 0.420 g of nitrogen monoxide with excess oxygen in the lab and produce 0.530 g of nitrogen dioxide. What is my percent yield for this reaction? (Hint: You have to write out the reaction to solve.)
82.3 %
4. Al + I
2
 AlI
3 a. What is the theoretical yield (g) if the reaction started with 1.20 g Al?
18.1 g AlI
3
(yield means product!) b. What is the percent yield if 30.0 g of AlI
3 is actually produced using the theor yield from a?
5. Al
2
S
82.9%
3
+ H
2
O  Al(OH)
3
3
+ H
2
S a. If 15.0 g of aluminum sulfide completely react what is the theor yield of aluminum hydroxide?
15.6 g Al(OH)

b. A student was in the doing the experiment above and collected 6.53g of Al(OH)
3
. What is her percent yield?
41.9%
6. Compound X is the product of a reaction. The reaction had a 75.2 % yield. Stoichiometric calculations predicted the product to have a mass of 23.5 grams. What was the actual yield?
17.7g
Set VII. Theoretical yield and percent yield. Worked Out
On a separate sheet of paper balance the equations for each problem and solve the questions.
1. Calculate the percent yield if the theoretical yield = 128 g, and actual yield = 112 g
87.5%
112g x 100 = 87.5%
128g
2. Calculate the actual yield if the theoretical yield = 23.98 kg, and the percent yield = 78.4%
18.8 kg = actual
3. I react 0.42 g of nitrogen monoxide with excess oxygen in the lab and produce 0.53 g of nitrogen dioxide.
What is my percent yield for this reaction? (Hint: You have to write out the reaction to solve.) 82.3 %
2NO + O
2
 2NO
2 to get the theoretical
0.42g NO x 1 mol NO x 2 mol NO
2
x 46.005g NO
2
1 30.006g NO 2 mol NO 1 mol NO
2
0.53 g NO
2
x 100 = 82.3 %
0.6439 g NO
2
= 0.6439g NO
2
= theoretical yield
4. 2 Al + 3 I
2
 2 AlI
3 a. What is the theoretical yield (g) if the reaction started with 1.20 g Al?
1.20 g Al x 1 mol Al x 2 mol AlI
3
x 407.64 g AlI
3
x 100 = 82.9% yield
= 18.1 g AlI3
1 26.982g Al 2 mol Al 1 mol AlI
3
b. What is the percent yield of 15 g of Al using the theor yield from a?
5. Al
2
S
15g AlI
3
18.1 g AlI
3
3
+ 6 H
2
O  2 Al(OH)
3
+ 3 H a. If 15 g of aluminum sulfide completely react what is the theor yield of aluminum hydroxide?
15g Al
2
S
2
S
3 x 1 mol Al
2
S
3
x 2 mol Al(OH)
3
1 150.162 g Al
2
S
3
1 mol Al
2
S
3
3
x 78.003g Al(OH)
3
= 15.6 g Al(OH)
1 mol Al(OH)
3 b. What is the percent yield of 6.53 g of aluminum hydroxide?
6.53 g Al(OH)
15.6 g Al(OH)
3
3
x 100 = 41.9%
6. Compound X is the product of a reaction. The reaction had a 75.2 % yield. Stoichiometric calculations predicted the product to have a mass of 23.5 grams. What was the actual yield?
75.2% = 100 x actual
23.5 g
17.7g

Objectives to be tested include, but are not exclusive to
1.
Interpret a chemical reaction in terms of atoms, molecules, moles
The coefficient of the substance will tell you the number of atoms, if there is an element that stands alone
(is. K NOT H
2
). The coefficient of a molecular compound will tell you the number or molecules of a compound. The coefficient of ANY substance will tell you the number of moles of that substance.
2.
Write mole relationships for any chemical reaction
First make sure the reaction is balanced. Once it is balanced, use the coefficients to determine the number of moles of that substance. Any substance can have a mole relationship or mole ratio as long as they are in the same balanced reaction.
3.
Determine moles, mass, molecules, or volume of any reactant or product given moles, mass, molecules or volume of any reactant or product
USE THE MOLE ROAD MAP
Moles: used a balanced reaction
Mass: calculate the molar mass
Volume: 22.4L = 1 mol of gas at STP
4.
Calculate % yield given info re actual and theoretical yields
Actual/theoretical *100 = % yield
5.
Define: stoichiometry, percent yield, actual yield, theoretical yield
Stoichiometry: The calculations or determination of quantities in chemical reactions
Percent yield: the ratio of actual yield to the theoretical yield multiplied by 100
Actual yield: the amount of product that actually formed when the reaction is carried out
Theoretical yield : maximum amount of product that could be formed from the balanced reaction
Practice
Consider the following reaction for questions 1 – 7.
4 HCl + O  2 H
1.
Complete the following table:
Substance
2 2
O + 2 Cl
Molar Mass (g/mol)
2
HCl
O
2
36.461
31.998
H
2
Cl
2
O 18.015
70.906
2.
How many mole relationships can be derived from this reaction? six
3.
Determine the mass of water produced from 10.5 moles of hydrochloric acid.
O 94.5g H
2
4.
Determine the number of moles of oxygen needed to make 53 L of chlorine gas.
(1 mol Cl
2
1.18mol O
= 22.4 L Cl
2
2
)
5.
Determine the mass of oxygen necessary to react with 32.3 g of HCl.

7.09 g O
2
6.
Determine the number of moles of chlorine gas produced from 17.2 moles of oxygen.
34.4 mol Cl
2
7.
Determine the missing yield for the following: a) actual yield = 53.9 g, theoretical yield = 65.4 g % yield = 82.4% b) actual yield = 0.56 g; percent yield = 42.1% theoretical yield = 1.33g
8.
Determine the percent yield for this reaction if 19.8 g Cl
2
is actually produced from 28.2 g of hydrochloric acid.
72.3%
9.
If 0.0468 mol of lead (II) nitrate are reacted with 0.065 mol of sodium chloride, then sodium nitrate and lead (II) chloride are formed. Which will be in excess? Calculate the mass of lead (II) chloride produced.
(Skeletal Equation: __ Pb(NO
3
)
2
+ __ NaCl  __ NaNO
3
+ __ PbCl
2
)
Lead nitrate in excess & sodium chloride is the limiting reactant. ~9.06g lead (II) chloride produced.

Objectives to be tested include, but are not exclusive to
 Interpret a chemical reaction in terms of atoms, molecules, moles
The coefficient of the substance will tell you the number of atoms, if there is an element that stands alone (is. K NOT H
2
). The coefficient of a molecular compound will tell you the number or molecules of a compound. The coefficient of ANY substance will tell you the number of moles of that substance.
 Write mole relationships for any chemical reaction
First make sure the reaction is balanced. Once it is balanced, use the coefficients to determine the number of moles of that substance. Any substance can have a mole relationship or mole ratio as long as they are in the same balanced reaction.
 Determine moles, mass, molecules, or volume of any reactant or product given moles, mass, molecules or volume of any reactant or product
Moles: used a balanced reaction
Mass: calculate the molar mass
Volume: 22.4L = 1 mol of gas at STP
 Calculate % yield given info re actual and theoretical yields
Actual/theoretical *100 = % yield
 Define: stoichiometry, percent yield, actual yield, theoretical yield
Stoichiometry: The calculations or determination of quantities in chemical reactions
Percent yield: the ratio of actual yield to the theoretical yield multiplied by 100
Actual yield: the amount of product that actually formed when the reaction is carried out
Theoretical yield : maximum amount of product that could be formed from the balanced reaction
Practice
Consider the following reaction for questions 1 – 7.
4 HCl + O  2 H
10.
Complete the following table:
Substance
2 2
O + 2 Cl
Molar Mass (g/mol)
2
HCl
O
2
36.461
31.998
H
2
Cl
2
O 18.015
70.906
11.
How many mole relationships can be derived from this reaction? six
12.
Determine the mass of water produced from 10.5 moles of hydrochloric acid.
10.5mol HCl | 2 mol H
| 4 mol HCl | 1 mol H
(1 mol Cl
53L Cl
2
2
2
O | 18g H
= 22.4 L Cl
| 1 mol Cl
2
2
)
| 1 mol O
2
| 22.4L Cl
2
| 2 mol Cl
2
2
O = 94.5g H
2
O
= 1.18mol O
2
2
O
13.
Determine the number of moles of oxygen needed to make 53 L of chlorine gas.

14.
Determine the mass of oxygen necessary to react with 32.3 g of HCl.
32.3g HCl | 1 mol HCl | 1 mol O
2
| 31.998g O
2
= 7.09 g O
2
| 36.461g HCl | 4 mol HCl | 1 mol O
2
15.
Determine the number of moles of chlorine gas produced from 17.2 moles of oxygen.
17.2 mol O
2
| 2 mol Cl
2
| 1 mol O
2
= 34.4 mol Cl
2
16.
Determine the missing yield for the following: a) actual yield = 53.9 g, theoretical yield = 65.4 g % yield = 82.4% b) actual yield = 0.56 g; percent yield = 42.1% theoretical yield = 1.33g
17.
Determine the percent yield for this reaction if 19.8 g Cl
2
is actually produced from 28.2 g of hydrochloric acid.
Theoretical: 28.2g HCl | 1 mol HCl | 2 mol Cl
2
| 70.906g Cl
| 36.461g HCl | 4 mol HCl | 1 mol Cl
2
2
= 27.4g Cl
2
% yield = actual
Theoretical
19.8g = 72.3%
27.4g
18.
If 0.0468 mol of lead (II) nitrate are reacted with 0.065 mol of sodium chloride, then sodium nitrate and lead (II) chloride are formed. Which will be in excess? Calculate the mass of lead (II) chloride produced.
Lead nitrate in excess & sodium chloride is the limiting reactant. ~9.06g lead (II) chloride produced.
Pb(NO
3
)
2
+ 2 NaCl  2NaNO
3
+ Pb(NO available
0.0468 mol Pb(NO
3
)
2
2 mol NaCl = 0.094 mol NaCl
1 mol Pb(NO
3
)
2 needed
3
)
2
0.065mol NaCl 1 mol Pb(NO
2 mol NaCl
3
)
2
= 0.0325 mol Pb(NO
3
) excess
2
Amount of lead (II) chloride produced:
0.065mol NaCl 1 mol PbCl
2
278.106g = 9.04g PbCl
2
2 mol NaCl 1 mol PbCl
2