BCLN CHEMISTRY 11 - Rev. July/2014
Unit 5 ~ Learning Guide Name: ______________________________
Instructions:
Using a pencil, complete the following notes as you work through the related lessons. Show ALL
work as is explained in the lessons. You are required to have this package completed BEFORE
you write your unit test. Do your best and ask questions if you don’t understand anything!
Chemical Equations:
1. What three things are conserved in a chemical reaction?
The total number of each kind of atom
The total mass
The total energy
2. The reaction which powers the engine in a car is as follows: 2 C6H18 + 2 1O2  12 CO2 + 18 H2O
A car burns 2000. g of octane gas (C8H18) and uses 7450. g of oxygen gas to do so. If this creates
5850. g of carbon dioxide then how many grams of water are created?
7450. g + 2000. g - 5850. g = 3600. g H2O
Balancing Equations:
1. What must you NEVER change when balancing an equation?
Subscripts
2. Fill in the following with the correct symbol.
Phase
Solid
Liquid
Gas
Aqueous
Symbol
(s)
(l)
(g)
(aq)
Page 1 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
3. Balance the following equations by filling in the blanks with the correct coefficient. Use the table
below each reaction to show your work. It is usually easiest to do the element found in the most
compounds LAST and to start with the element that is simplest to balance. An example is
provided for you. where H was balanced LAST and Al was balanced first.
_1__Al4C3 + _12_H2O  __3_CH4 + _4__Al(OH)3
Reactants
Products
Al 4
Al 1x4 = 4
C3
C 1x3 =3
H 2x12 = 24
H 4x3 =12 + 3x4 = 12 = 24
O 1x12 = 12
O 3x4 = 12
a) __1_Mg + __2_HCl  __1_MgCl2 + __1_H2
Reactants
Products
Mg 1
Mg 1
H 1x2 = 2
H2
Cl 1x2 = 2
Cl 2
b) __2_BiCl3
+ __3_H2S  _1_Bi2S3 + __6_HCl
Reactants
Products
Bi 1x 2 = 2
Bi 2
Cl 3 x 2 = 6
Cl 1 x 6 = 6
H 2x3 = 6
H 1x6=6
S 1x3 = 3
S3
c) __3_Fe
+ __2_O2  _1_Fe3O4
Reactants
Products
Fe 1x3 = 3
Fe 3
O 2x2=4
O4
Page 2 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
d) __2_H2O2
 _2__H2O + __1_O2
Reactants
Products
H 2x2 = 4
H 2x2 = 4
O 2x2=4
O 1x2 =2 + 2 =
4
e) _2_K2CrO4 + _2__HCl  __1_K2Cr2O7 + __2_KCl + __1_H2O
Reactants
Products
K 2x2 = 4
K2
Cr 1x2 = 2
Cr 2
O 4x2 = 8
O7
H 1x2 =2
H2
Cl 1x2 = 2
Cl 1x2 = 2
f) __1_C6H12O6
+ 1x2 = 2 = 4
+
1
= 8
+ _6__O2  __6_CO2 + _6__H2O
Reactants
Products
C6
C 1x6 = 6
H 12
H 2x6 = 12
O 6 + 2x6 = 12 = 18
O 2x6 = 12 + 1x6 = 6 = 18
g ) _2__C8H18 + _25_O2  _16__CO2 + _18__H2O
Reactants
Products
C 8x2 = 16
C 1x16 = 16
H 18x2 = 36
H 2x18 = 36
O 2x25 = 50
O 2x16 = 32 + 1x18 = 18 = 50
Page 3 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
Types of Chemical Reactions:
1. What are the 6 types of reactions you are expected to be able to recognize in Chemistry 11?
Combination (Synthesis), Decomposition, Single Replacement, Double Replacement,
Combustion, Acid-Base (Neutralization)
2. Please classify the following reactions as one of the 6 types you have learned about. IF you want
extra practice see if you can balance them as well. You may use the following abbreviations of
the types: Syn, Dec, SR, DR, Comb, Neut.
a) ________Dec______
_2_ H2 O2  _2_ H2O + _1_O2
b) _______Neut______
_1_HCl + _1_ NaOH  _1_NaCl +_1_H2O
c) _________SR______
_2_Al + _3_ NiBr2  _2_AlBr3 + _3_Ni
d) ________Syn______
_1_N2 + _3_ H2  _2_ NH3
e) _________DR_____
_1_AgNO3 + _1_ KCl  _1_ AgCl + _1_ KNO3
f) _______Comb_____
_2_ C6H6 + _15_ O2  _12_ CO2 + _6_ H2O
g) _________SR_____
_1_Mg + _2_ HCl  _1_ MgCl2 + _1_ H2
h) ________Syn______
_3_ Mg + _1_ N2  _1_ Mg3N2
i) _________SR______
_1_ Ba + _2_ HBr  _1_ BaBr2 + _1_ H2
j) _________DR______
_2_ BiCl3 + _3_ H2S  _1_ Bi2S3 + _6_ HCl
k) _________SR______
_1_ Br2 + _2_ KI  _1_ I2 + _2_ KBr
l) _________Syn_____
_3_ Fe + _2_ O2  _1_ Fe3O4
m) _______Comb_____
_2_ C4H10 + _13_ O2  _8_ CO2 + _10_ H2O
n) _________SR______
_2_ CaO + _4_ C  _2_ CaC2 + _1_ O2
o) ________Neut_____
_1_ LiOH + _1_ HBr  _1_ LiBr + _1_ H2O
p) _________SR______
_1_ Bi2O3 + _3_ H2  _2_ Bi + _3_ H2O
q) ________Syn______
_1_ P4 + _6_ I2  _4_ PI3
Page 4 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
r) _______SR_______
_2_ Fe2O3 + _3_ C  _4_ Fe + _3_ CO2
s) _______Comb_____
_1_ C6H12O6 + _6_ O2  _6_ CO2 + _6_ H2O
t) _______Dec_______
_1_ NH4NO3  _1_ N2O + _2_ H2O
u) _______DR_______
_1_ Al4C3 + _12_ H2O  _3_ CH4 + _4_ Al(OH)3
v) _______Syn_______
_1_ Ca(OH)2 + _2_ CO2  _1_ Ca(HCO3)2
Note that balancing is optional for students on this question.
Energy:
1. State whether each of the following are exothermic or endothermic.
Reaction or Event
Exo or Endo
2 C8H18 + 25 O2  16 CO2 + 16 H2O + 10110 kJ
Exo
Ba(OH)2 + 2 NH4Cl + 430 kJ 
BaCl2 + 2 NH4OH
6SOCl2 + CoCl26H2O  CoCl2 + 12HCl + 6SO2
Mn(s) + 2 HCl(aq)  MnCl2(aq) + H2(g)
H = +360 kJ
H = -221 kJ
Endo
Endo
Exo
Boiling Water
Endo
Metabolizing Food
Exo
The energy possessed by the products is greater than the energy
possessed by the reactants.
Endo
2. How much heat is evolved when 30.g of magnesium is burned in excess oxygen gas given the
reaction below:
2Mg + O2  2 MgO
𝟑𝟎. 𝐠 𝐌𝐠 𝐱
𝟏 𝐦𝐨𝐥 𝐌𝐠
𝟐𝟒.𝟑𝟏 𝐠 𝐌𝐠
H = -600. kJ
𝐱
𝟔𝟎𝟎 𝐤𝐉
𝟐 𝐦𝐨𝐥 𝐌𝐠
= 𝟑𝟕𝟎 𝐤𝐉
Page 5 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
Stoichiometry:
1. Use the balanced equation below to answer the following questions.
1 Mg + 2 HCl  1 MgCl2 + 1 H2
a) How many moles of hydrogen gas are produced when 10 moles of hydrochloric acid
completely react?
𝟏𝟎 𝐦𝐨𝐥 𝐇𝐂𝐥 𝐱
𝟏 𝐦𝐨𝐥 𝐇𝟐
= 𝟓 𝐦𝐨𝐥 𝐇𝟐
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
b) What mass of magnesium chloride is produced if 5.75 mol of magnesium reacts completely?
Molar Mass of MgCl2 = 1 Mg + 2 Cl = 1(24.31) + 2(35.45) = 95.21 g/mol
𝟓. 𝟕𝟓 𝐦𝐨𝐥 𝐌𝐠 𝐱
𝟏 𝐦𝐨𝐥 𝐌𝐠𝐂𝐥𝟐
𝟏 𝐦𝐨𝐥 𝐌𝐠
𝐱
𝟗𝟓.𝟐𝟏𝐠 𝐌𝐠𝐂𝐥𝟐
𝟏 𝐦𝐨𝐥 𝐌𝐠𝐂𝐥𝟐
= 𝟓𝟒𝟕𝐠 𝐌𝐠𝐂𝐥𝟐
c) What mass of magnesium is needed to react 15.2g of hydrochloric acid?
Molar Mass of HCl = 1 H + 1 Cl = 1(1.01) + 1(35.45) = 36.46 g/mol
𝟏𝟓. 𝟐𝐠 𝐇𝐂𝐥 𝐱
𝟏 𝐦𝐨𝐥 𝐇𝐂𝐥
𝟑𝟔.𝟒𝟔𝐠 𝐇𝐂𝐥
𝐱
𝟏 𝐦𝐨𝐥 𝐌𝐠
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
𝐱
𝟐𝟒.𝟑𝟏 𝐠 𝐌𝐠
𝟏 𝐦𝐨𝐥 𝐌𝐠
= 𝟓. 𝟎𝟕𝐠 𝐌𝐠
d) What mass of magnesium is needed to create 87.5 L of hydrogen gas at SATP?
𝟖𝟕. 𝟓𝐋 𝐇𝟐 𝐱
𝟏 𝐦𝐨𝐥 𝐇𝟐
𝟐𝟒.𝟖𝐋 𝐇𝟐
𝐱
𝟏 𝐦𝐨𝐥 𝐌𝐠
𝟏 𝐦𝐨𝐥 𝐇𝟐
𝐱
𝟐𝟒.𝟑𝟏 𝐠 𝐌𝐠
𝟏 𝐦𝐨𝐥 𝐌𝐠
= 𝟖𝟓. 𝟖𝐠 𝐌𝐠
e) How many molecules of hydrogen gas are created if 6.6 mol of hydrochloric acid react with
3.3 mol of magnesium?
𝟔. 𝟔 𝐦𝐨𝐥 𝐇𝐂𝐥 𝐱
𝟏 𝐦𝐨𝐥 𝐇𝟐
𝟐 𝐦𝐨𝐥 𝐇𝐂𝐥
𝐱
𝟔.𝟎𝟐𝐱𝟏𝟎𝟐𝟑 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞𝐬 𝐇𝟐
𝟏 𝐦𝐨𝐥 𝐇𝟐
Page 6 of 12
= 𝟐. 𝟎𝐱𝟏𝟎𝟐𝟒 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞𝐬 𝐇𝟐
BCLN CHEMISTRY 11 - Rev. July/2014
2. Use the balanced equation below to answer the following questions.
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(l) + 10110 kJ
a) How many moles of water are created when 108 moles of oxygen gas completely react?
𝟏𝟎𝟖 𝐦𝐨𝐥 𝐎𝟐 𝐱
𝟏𝟖 𝐦𝐨𝐥 𝐇𝟐 𝐎
= 𝟕𝟕. 𝟖 𝐦𝐨𝐥 𝐇𝟐 𝐎
𝟐𝟓 𝐦𝐨𝐥 𝐎𝟐
b) How many litres of carbon dioxide gas will be created with 525 L of oxygen gas react
completely at STP? Hint this can be done with a single conversion!
𝟓𝟐𝟓 𝐋 𝐎𝟐 𝐱
𝟏𝟔 𝐋 𝐂𝐎𝟐
𝟐𝟓 𝐋 𝐎𝟐
= 𝟑𝟑𝟔𝐋 𝐂𝐎𝟐
c) How much energy is created if a sample of C8H18 containing 7.63 x 1025 atoms of hydrogen
reacts completely?
𝟕. 𝟔𝟑𝐱𝟏𝟎𝟐𝟓 𝐚𝐭𝐨𝐦𝐬 𝐇 𝐱
𝟏 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞 𝐂𝟖 𝐇𝟏𝟖
𝟏𝟖 𝐚𝐭𝐨𝐦𝐬 𝐇
𝐱
𝟏 𝐦𝐨𝐥 𝐂𝟖 𝐇𝟏𝟖
𝟔.𝟎𝟐𝐱𝟏𝟎𝟐𝟑 𝐦𝐨𝐥𝐞𝐜𝐮𝐥𝐞 𝐂𝟖 𝐇𝟏𝟖
𝐱
𝟏𝟎𝟏𝟏𝟎 𝐤𝐉
𝟐 𝐦𝐨𝐥 𝐂𝟖 𝐇𝟏𝟖
= 𝟑. 𝟓𝟔𝐱𝟏𝟎𝟒 𝐤𝐉
d) What volume of carbon dioxide gas at SATP is created in a reaction which produces
5.37x106 kJ of energy?
𝟓. 𝟑𝟕𝐱𝟏𝟎𝟔 𝐤𝐉 𝐱
𝟏𝟔 𝐦𝐨𝐥 𝐂𝐎𝟐
𝟏𝟎𝟏𝟏𝟎 𝐤𝐉
𝐱
𝟐𝟒.𝟖 𝐋 𝐂𝐎𝟐
𝟏 𝐦𝐨𝐥 𝐂𝐎𝟐
= 2.11x𝟏𝟎𝟓 𝐋 𝐂𝐎𝟐
e) What mass of octane gas (C8H18) is required to create 112 g of water?
Molar Mass C8H18 = 8 C + 18 H = 8(12.01) + 18(1.01) = 114.26 g/mol
Molar Mass H2O = 2 H + 1 O = 2(1.01) + 16.00 = 18.02 g/mol
𝟏𝟏𝟐𝐠 𝐇𝟐 𝐎 𝐱
𝟏 𝐦𝐨𝐥 𝐇𝟐 𝐎
𝟏𝟖.𝟎𝟐𝐠 𝐇𝟐 𝐎
𝐱
𝟐 𝐦𝐨𝐥 𝐂𝟖 𝐇𝟏𝟖
𝟏𝟖 𝐦𝐨𝐥 𝐇𝟐 𝐎
𝐱
𝟏𝟏𝟒.𝟐𝟔𝐠 𝐂𝟖 𝐇𝟏𝟖
𝟏 𝐦𝐨𝐥 𝐂𝟖 𝐇𝟏𝟖
Page 7 of 12
= 𝟕𝟖. 𝟗𝐠 𝐂𝟖 𝐇𝟏𝟖
BCLN CHEMISTRY 11 - Rev. July/2014
3. To do each of these questions you will need the result from the previous question. The answers
for each are provided but please do your best to try and do them on your own before checking the
answers.
a) If barium hydroxide reacts with ammonium nitrate what type of reaction is occurring?
double replacement
b) What are the names of the two products of this reaction?
barium nitrate and ammonium hydroxide
c) Write and then balance the equation for this reaction. Remember you need to make sure to
correctly balance each ion with each other to create the correct chemical formulas.
Ba(OH)2 + 2 NH4NO3  Ba(NO3)2 + 2 NH4OH
d) What mass of ammonium hydroxide is created if 26.4g of barium hydroxide reacts
completely?
Molar Mass Ba(OH)2 = 1 Ba + 2 O + 2 H = 1(137.33) + 2(16.00) + 2(1.01) = 171.35 g/mol
Molar Mass NH4OH = 1 N + 5H + 1 O = 1(14.01) + 5(1.01) + 1(16.00) = 35.06 g/mol
𝟐𝟔. 𝟒𝐠 𝐁𝐚(𝐎𝐇)𝟐 𝐱
𝟏 𝐦𝐨𝐥 𝐁𝐚(𝐎𝐇)𝟐
𝟏𝟕𝟏.𝟑𝟓𝐠 𝐁𝐚(𝐎𝐇)𝟐
𝐱
𝟐 𝐦𝐨𝐥 𝐍𝐇𝟒 𝐎𝐇
𝟏 𝐦𝐨𝐥 𝐁𝐚(𝐎𝐇)𝟐
𝐱
𝟑𝟓.𝟎𝟔𝐠 𝐍𝐇𝟒 𝐎𝐇
𝟏 𝐦𝐨𝐥 𝐍𝐇𝟒 𝐎𝐇
= 𝟏𝟎. 𝟖𝐠 𝐍𝐇𝟒 𝐎𝐇
e) If 37.4 kJ of energy are required for every 2 moles of ammonium nitrate that react how much
energy is required to cause 48.6g of ammonium nitrate to react completely?
Molar Mass NH4 NO3 = 2 N + 4H + 3 O = 2(14.01) + 4(1.01) + 3(16.00) = 80.06 g/mol
𝟒𝟖. 𝟔𝐠 𝐍𝐇𝟒 𝐍𝐎𝟑 𝐱
𝟏 𝐦𝐨𝐥 𝐍𝐇𝟒 𝐍𝐎𝟑
𝟖𝟎.𝟎𝟔𝐠 𝐍𝐇𝟒 𝐍𝐎𝟑
𝐱
𝟑𝟕.𝟒 𝐤𝐉
𝟐 𝐦𝐨𝐥 𝐍𝐇𝟒 𝐍𝐎𝟑
Page 8 of 12
= 𝟏𝟏. 𝟒 𝐤𝐉
BCLN CHEMISTRY 11 - Rev. July/2014
Limiting Reagents:
1. Define the following terms and provide the formula for the third.
a) Limiting Reagent - The reactant in a reaction that runs out or is totally consumed
b) Excess Reagent - The reactant in a reaction that remains after the reaction is complete
c) Percent Yield - a measure of how much product was actually obtained in an experiment
compared to how much was predicted to be made
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐘𝐢𝐞𝐥𝐝 =
𝐀𝐜𝐭𝐮𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
𝐱𝟏𝟎𝟎%
2. Consider the following reaction: 2Al + 3 NiBr2  2 AlBr3 + 3 Ni.
27.0 g of aluminum react with 219 g nickel (II) bromide.
a) Which reactant is the limiting reagent? Show your work!
1. Moles of Al present:
27.0 g Al 
1 mol Al
 1.00 mol Al is present
26.98 g Al
2. Moles of NiBr2 present:
Molar Mass of NiBr2 = 1 Ni + 2 Br = 1(58.69) + 2(79.90) = 218.49 g/mol
219 g NiBr2 
1 mol NiBr2
 1.00 mol NiBr2 is present
218.49 g NiBr2
3. Moles of NiBr2 required to react with all the Al present:
1.00 mol Al 
3 mol NiBr2
 1.50 mol NiBr2 is required
2 mol Al
4. Because Moles NiBr2 present < Moles of NiBr2 required, NiBr2 is the Limiting Reagent
Page 9 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
b) What mass of excess reagent is left over when the reaction is complete
Moles
Excess Reagent
Limiting Reagent
Product
Product
2Al
3NiBr2
2AlBr3
3Ni
initial moles
change in moles
final moles
1.00
- 0.667
0.333
1.00
- 1.00
0
0
+0.667
0.667
0
+1.00
1.00
× 2/3
× 2/3
Mass of Excess Reagent (Al) left over: 0.333 mol Al 
26.98 g Al
 8.98 g Al
1 mol Al
c) What mass of aluminum bromide is created in the reaction? Be sure to use the limiting
reagent in this calculation
Molar Mass of AlBr3 = 1 Al + 3 Br = 1(26.98) + 3(79.90) = 266.68 g/mol
Mass of AlBr3 created = 0.667 mol AlBr3 
266.68 g AlBr3
 178 g AlBr3
1 mol AlBr3
3. Answer these questions based on this reaction and the following information
Mn(s) + 2 HCl(aq)  MnCl2(aq) + H2(g)
H = -221 kJ
115g of manganese is placed in a 1.25 L of a 3.50 M solution of hydrochloric acid.
a) Which reactant is the limiting reagent? Show your work!
1. Moles of Mn present:
115 g Mn 
1 mol Mn
 2.093 mol Mn is present
54.94 g Mn
2. Moles of HCl present:
3.50 mol HCl
 1.25 L  4.375 mol HCl is present
1L
3. Moles of HCl required to react with all the Mn present:
2.093 mol Mn 
2 mol HCl
 4.186 mol HCl is required
1 mol Mn
4. Because Moles of HCl present > Moles of HCl required, HCl is the Excess Reagent.
Therefore, Mn is the Limiting Reagent.
Page 10 of 12
BCLN CHEMISTRY 11 - Rev. July/2014
b) How many moles of excess reagent is left over when the reaction is complete?
Moles
Limiting Reagent
Excess Reagent
Mn
2HCl
Product
Product
MnCl2
H2
initial moles
2.093 × 2/1
4.375
0
change
in
moles
2.093
4.186
+2.093
c) What is the theoretical mass of hydrogen gas created in the reaction?
final moles
0
0.189
2.093
0
+2.093
2.093
Moles of Excess Reagent (HCl) Left over = 0.189 moles
c) What is the theoretical mass of hydrogen gas created in the reaction?
Theoretical Mass of H2(g): 2.093 mol H 2 
2.02 g H 2
 4.23 g H 2
1 mol H 2
d) If 33.6 L of hydrogen gas was measured to have formed at STP what is the percent yield of this
reaction? Hint: You will need to convert this amount to grams!
𝐀𝐜𝐭𝐮𝐚𝐥 𝐘𝐢𝐞𝐥𝐝 = 𝟑𝟑. 𝟔𝐋 𝐇𝟐 𝐱
𝐏𝐞𝐫𝐜𝐞𝐧𝐭 𝐘𝐢𝐞𝐥𝐝 =
𝟏 𝐦𝐨𝐥 𝐇𝟐
𝟐𝟐.𝟒 𝐋 𝐇𝟐
𝐀𝐜𝐭𝐮𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐘𝐢𝐞𝐥𝐝
𝐱
𝟐.𝟎𝟐𝐠 𝐇𝟐
𝟏 𝐦𝐨𝐥 𝐇𝟐
𝐱𝟏𝟎𝟎% =
Answers:
Chemical Equations:
2. 3600. g H2O
Balancing Equations:
3. a) 1,2,1,1
e) 2, 2,1,2,1
b) 2,3,1,6
c) 3,2,1
f) 1,6,6,6
g) 2,25,16,18
d) 2, 2, 1
Page 11 of 12
= 𝟑. 𝟎𝟑𝐠 𝐇𝟐
𝟑.𝟎𝟑𝐠 𝐇𝟐
𝟒.𝟐𝟑𝐠 𝐇𝟐
𝐱𝟏𝟎𝟎% = 𝟕𝟏. 𝟔%
BCLN CHEMISTRY 11 - Rev. July/2014
Energy
2. 370 kJ
Stoichiometry
1. a) 5 mol H2
b) 547g MgCl2
c) 5.07g Mg
e) 2.0 x 1024 molecules H2
d) 85.8g Mg
2. a) 77.8 mol H2O
b) 336 L CO2
d) 2.11 x 105 L CO2
3. a) double replacement
c) 3.56 x 104 kJ
e) 78.9g C8H18
b) barium nitrate and ammonium hydroxide
c) Ba(OH)2 + 2 NH4NO3  Ba(NO3)2 + 2 NH4OH
d) 10.8 g NH4OH
e) 11.4 kJ
Limiting Reagents:
2. a) NiBr2
b) 8.98 g Al
c) 178 g AlBr3
3. a) Mn
b) 0.189 mol HCl
c) 4.23 g H2
Page 12 of 12
d) 71.6%