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Ch.7 Chemical Reactions and Stoichiometry, ㅤ4 hours in the working -5PM …
Ch.7 Chemical Reactions
and Stoichiometry
Ch 7.2 Limiting and Excess Reactants
What is Limiting and
Excess Reactants?
Limiting Reactants:
When there isn’t too much to one reacted so it is fully used after the reaction.
Excess Reactants:
When there is an extra of a reaction so after the reaction it's still there.
How to Identifying
the Limiting Reactants?
Requirements...
Balanced the equation
The moles of the Reactants if the mass is given use molar mass to get moles
Now you have to do the mole ratio with any of the produced and the lower amount will be the
Limiting Reactant
How to find moles if the mass is given?
Let's say I need to find the moles of O2 and the I know that the mass of the oxygen is 10g
Formula: mol = mass ÷ molar mass
molar mass of O2 = 16g/mol x 2 =
32 g/mol
mol = 10g ÷ 32 g/mol ≈
0.31 mol
of O2
How to Determine Amount
of Product Formed?
Requirements...
Balanced the equation
Identify the
Limiting Reactant
Use the mole ratio between the
Limiting Reactant
and the
product
Limiting Reactant question
Step 1 find the molar masses
molar mass of N2 = 14g/mol x 2 =
28g/mol
molar mass of H2 = 1g/mol x 2 =
2g/mol
Step 2 using the molar masses & mass, find moles
moles of N2 = 28g ÷ 28g/mol =
1 mol
moles of H2 = 2g ÷ 12g/mol =
0.16... mol
Step 3 mole ratio between the reactants (N&H) and NH3
N2 ㅤ + ㅤ 3H2 ㅤ --> ㅤ ㅤ 2NH3
1 mol---------------------->
ㅤ2 mol E.R.
ㅤㅤㅤㅤ 0.16 mol ---->
0.106 mol L.R.
Answer:
H2 is the Limiting Reactant
There are
28g of N2
and
12g of H2
which one is the Limiting Reactant?
Balanced Equation:
N2 + 3H2 --> 2NH3
Finding moles of product
Step 1 Identify the equation has given
L.R. is hydrogen
and
0.16 mol
of H2
An excess of nitrogen reacts with
0.16 moles of hydrogen
how many grams of
NH3 will be produced?
Balanced Equation:
N2 + 3H2 --> 2NH3
Step 2 Identify what is required
The
mass of NH3
is needed
Step 3 mole ratio between the reactant (H) and NH3
The coefficients are on the top
The moles are on the bottom
ㅤㅤㅤㅤ 2ㅤ ㅤㅤㅤㅤㅤㅤㅤㅤㅤ ㅤㅤㅤ3 ㅤ
----------------------------- ㅤ:ㅤ---------------------------
ㅤㅤㅤㅤ xㅤㅤㅤㅤㅤ ㅤㅤㅤ 0.16 mol of H ㅤ
x =
0.106 mol of NH3
Step 4 find the molar mass of NH3
molar mass of NH3 = 14g/mol + 3 x 1g/mol =
17g/mol
Step 5 find mass using moles and molar mass
Using the formula
mol = mass ÷ molar mass
, it can be rearranged to
mass = mol x molar mass
,
0.106 mol x 17g/mol =
1.8g
Answer: 1.8g of NH3 is produced
Ch 7.1 Stoichiometry
Particle Ratios in a Balanced Chemical Equation
The
Coefficient
in a balanced equation shows the number of particles in the chemical reaction
If you wanted
10 molecules of Ammonia
ㅤ
how much
nitrogen molecule do you need?
N2 + 3H2 --> 2NH3
xㅤ ㅤㅤㅤ ㅤㅤㅤ1 molecule of N2ㅤ
------------------------------ ㅤ=ㅤ---------------------------
10 molecules of NH3 ㅤㅤㅤ 2 molecules of NH3
x =
5 molecules of N2
What is Stoichiometry?
Stoichiometry:
The study of the number of reactants and products in a chemical reaction.
Stoichiometric Amount:
When the molar amount of a reactant or product is dependent on a blanched chemical equation.
Stoichiometry
is so important because just a strong understanding of the subject could save a lot of excessive reactants and products so everything thing could as effective as possible when it comes to reacting substances.
Mole Ratios in a Balanced
Chemical Equation
The coefficients
are on this side
n NH3ㅤ ㅤㅤㅤ ㅤㅤㅤ2 mol of NH3 ㅤ
------------------------------ ㅤ=ㅤ---------------------------
9 mol of hydrongen ㅤㅤㅤㅤㅤ 3 mols of N2 ㅤ
this side is called
"Mole Ratio"
ㅤ
n = moles =
6 moles of NH3
Let's say an excess of nitrogen reacts with
9 moles of hydrogen
, how many moles of
NH3 will be produced?
Blanced equation: N2 + 3H2 --> 2NH3
The only condition is the
equation has to be balanced
to do the mole ratio
Mole Ratio:
The ratio of 2 substances in a balanced chemical equation.
Mass Relationships in Chemical Equation
Using the equation:
m = n x M
gives the mass
where
m = mass in g
n = mol
M = g/mol / molar mass
What is the
mass of N
in this equation
N2 + 3H2 --> 2NH3?
n ㅤ×ㅤㅤM ㅤ= ㅤ m
1 mol × 14 g/mol = 14 g
14g of Nitrogen
Ch 7.3 Reaction Yields
What is Reaction Yields
Theoretical Yield:
The amount of product predicted from using stoichiometric calculations.
Actual Yield:
The actual real-life results.
In the real experiment, the Theoretical Yield will most likely not match the Actual and their are multiply reason why.
Factors Affecting Actual Yield
competing reaction
this is when there is another reaction is taking place for example let's say a reaction produces CO2 when it's going through complete combustion but when there is a low amount of oxygen an competing reaction produces CO
Impure
, the substance itself may not be 100% pure like when u have cheap gold only a % of it is gold
Reaction rates
this is when the reaction happens too slowly and this cause it not to complete the reaction by the time it is measured also Reaction is affected by surface area, temperature, pressure, and reaction vessel conditions
Laboratory techniques
, this is when of the product get stuck one the sides of the equipment like, in a water bottle theirs always water on the side stuck there
Calculating the Percentage Yield
Percentage Yield
is the percentage of the Actual Yield over the Theoretical Yield.
Requirements...
Balanced the equation
The amount of a
reactant
The
actual yield
of a
product
Step 1 identity what is given
20g of N2, 2g NH3 actual yield and N is the L.R.
When
20g of nitrogen
reacts with
sufficient hydrogen
. If
2g of ammonia
is
obtained
by experiment, what is the percentage yield of the reaction?
Balanced Equation:
N2 + 3H2 --> 2NH3
Step 4 find moles of nitrogen
mol = mass ÷ molar mass
20g ÷ 28g/mol = 0.71mol
Step 2 identity what is required
The
percentage yield
of NH3
Step 3 find molar masses
molar mass of NH3 = 14g/mol + 1 x 3g/mol =
17g/mol
molar mass of N2 = 2 x 14g/mol =
28g/mol
Step 5 mole ratio between N2 and NH3
1:2 = N2:NH3 = 0.71 mol of N2 :
1.42 mol of NH3
Step 7 find the Percentage Yield
Actual Yield over the Theoretical Yield
2g ÷ 24.14g x 100% =
8.3%
Answer:
The percentage Yield of NH3 is 8.3%
Step 6 use moles and molar mass to find the mass of NH3
mass = mol x molar mass 1.42 mol x 17g/mol =
24.14g of NH3
The importance of the Actual Percentage Yield
It's critical in order to get maximum efficiency out of the products
It's important to understand why maximum yield wasn't reached
If the percentage yield is known, the amount of reactants can be easily calculated which means nothing should be wasted
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ㅤ2022-6-10
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