IN OUR PREVIOUS CLASS ABOUT CHEMICAL
EQUATIONS, WE FOUND THAT AN EQUATION
CAN BE USED TO DESCRIBE A CHEMICAL
REACTION. IT GIVES US A GREAT DEAL OF
INFORMATION:
CH4
+
2O2

CO2
+
2H2O
1 MOLECULE 2 MOLECULES
METHANE
OXYGEN
1 MOLECULE
2 MOLECULES
CARBON DIOXIDE WATER
1 MOLE
2 MOLES
1 MOLE
2 MOLES
16 g
64 g
44 g
36 g
6.02 x 1023
Molecules
2(6.02x1023)
molecules
6.02 x 1023
molecules
2(6.02 x 1023)
molecules
THE MOST USEFUL RELATIONSHIIPS WILL
BE THE MOLAR RATIOS INVOLVED IN THE
REACTIONS AND THE MASS RATIOS.
WE CAN USE THIS INFORMATION TO
CALCULATE:
1. THE AMOUNTS OF STARTING MATERIALS
NEEDED TO MAKE A GIVEN AMOUNT OF
PRODUCT
2. THE THEORETICAL YIELD (100%) THAT
GIVEN AMOUNTS OF PRODUCTS WOULD
GIVE
3. THE LIMITING REAGENT IN A REACTION
CONSIDER THE FOLLOWING PROBLEM.
HOW MANY MOLES OF OXYGEN CAN YOU
PRODUCE FROM 5 MOLES OF KClO3?
5 moles
2KClO3  2KCl
2 moles
+
X moles
3O2
3 moles
THE EQUATION TELLS YOU THAT 2 moles
OF KClO3 WILL GIVE YOU 3 moles OF O2.
YOU HAVE 5 MOLES AND NEED TO FIND
HOW MUCH O2 WILL BE PRODUCED.
USE THE RATIO: 5/2 = X/3
X = 3(5/2) = 7.5 moles
AND:
YOU CAN ALSO USE YOUR EQUATIONS IN
TERMS OF MASS.
HOW MANY GRAMS OF WATER ARE PRODUCED
IN THE OXIDATION OF 1 GRAM OF GLUCOSE?
C6H12O6 + 6O2  6CO2 + 6H2O
(combustion reaction)
1 MOLE
GLUCOSE
6 MOLES
WATER
LET’S CALCULATE THE MOLAR MASSES OF
WATER AND GLUCOSE SO WE CAN GET THE
MASS RELATIONSHIPS.
The molar mass of glucose is:
6 C = 6 x 12.0 =
72.0 g
12 H = 12 x 1.01 = 12.12 g
6 O = 6 x 16.0 =
96.0 g
molar mass
= 180.0 g/mole
The molar mass of
2 H = 2 x 1.01
1 ) = 1 x 16.0
molar mass
water is:
= 2.02 g
= 16.0 g
= 18.02 g/mole
1.0 g glucose
X g water
C6H12O6 + 6O2  6CO2 + 6H2O
180 g/mole
6(18 g/mole)
1/180 = X/108
OR
X = 108/180 = 0.6 grams WATER
The decomposition of KClO3 (potassium
chlorate) is commonly used to prepare small
amounts of O2 in the laboratory:
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
How many grams of O2 can be prepared from
4.50 g KClO3?
FIRST, THE FORMULA MASS OF KCLO3 IS:
1 K = 1 x 39.1 = 39.1 g
1 Cl = 1 x 35.5 = 35.5 g
3 O = 3 x 16.0 = 48.0 g
formula mass = 123 g/mole
THE FORMULA MASS OF O2 IS 2 X 16 = 32 g/mole
4.5 g
2 KClO3

2(123 g/mole)
2 KCl
+
X g
3 O2
3(32 g/mole)
SO, YOUR RATION IS:
4.5/246 = X/96
AND
X = (4.5/246) x 96 = 1.76 g OXYGEN
2 KClO3 (s)  2 KCl (s) + 3 O2 (g)
SOLID LITHIUM HYDROXIDE IS USED IN
SPACE VEHICLES TO REMOVE EXHALED
CARBON DIOXIDE. THE LITHIUM
HYDROXIDE REACTS WITH GASEOUS
CARBON DIOXIDE TO FORM SOLID LITHIUM
CARBONATE AND LIQUID WATER. HOW
MANY GRAMS OF CARBON DIOXIDE CAN BE
ABSORBED BY 1.00 G OF LITHIUM
HYDROXIDE?
YOU NEED THE BALANCED EQUATION, SO
2 LiOH (s) + CO2  Li2CO3 (s) + H2O (l)
The formula mass of LiOH is:
1 Li = 1 x 6.94 = 6.94 g
1 O = 1 x 16.0 = 16.0 g
1 H = 1 x 1.01 = 1.01 g
formula mass = 24.0 g and you have 2, so 2
x 24.0 = 48.0 g
The molar mass of CO2 = 12.0 + 2 x 16.0 = 44.0 g
1.0 g
X g
2 LiOH (s) + CO2  Li2CO3 (s) + H2O (l)
48 g
44.0 g
So,
44 / 48 = X / 1.00 and X = (44/48) x 100
X = 0.917 g
http://www.gpb.org/chemistry-physics/chemistry/801