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Lecture #9
Calculating Amounts of Reactants
and Products
and
The Limiting Reactant Concept
Chemistry 142 B
James B. Callis, Instructor
Autumn Quarter, 2004
Chemical Equation Calculations
Mass
Atoms (Molecules)
Avogadro’s
Number
Reactants
6.02 x 1023
Molecules
Moles
Molecular
g/mol
Weight
Products
Problem 9-1: Calculating Reactants and
Products in a Chemical Reaction - I
Problem: Given the following chemical reaction between aluminum
sulfide and water, if we are given 65.80 g of Al2S3: a) How many moles
of water are required for the reaction? b) What mass of H2S & Al(OH)3
would be formed?
Al2S3 (s) + 6 H2O (l)
2 Al(OH)3 (s) + 3 H2S (g)
Plan: Calculate moles of aluminum sulfide using its molar mass, then
from the equation, calculate the moles of water, and then the moles of
hydrogen sulfide, and finally the mass of hydrogen sulfide using
it’s molecular weight.
Solution:
a) molar mass of aluminum sulfide =
moles Al2S3 =
Calculating Reactants and Products in a
Chemical Reaction - II
a) cont.
H2O: 0.4382 moles Al2S3 x ___ moles H2O
1 mole Al2S3
b)H2S: 0.4382 moles Al2S3 x ___ moles H2S
1 mole Al2S3
molar mass of H2S =
mass H2S =
Al(OH)3: 0.4382 moles Al2S3 x
molar mass of Al(OH)3 =
mass Al(OH)3 =
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - I
Problem 9-2: Calcium Phosphate could be prepared in the following
reaction sequence:
4 P4 (s) + 10 KClO3 (s)
P4O10 (s) + 6 H2O (l)
2 H3PO4 (aq) + 3 Ca(OH)2 (aq)
4 P4O10 (s) + 10 KCl (s)
4 H3PO4 (aq)
6 H2O(aq) + Ca3(PO4)2 (s)
Given: 15.5 g P4 and sufficient KClO3 , H2O and Ca(OH)2. What mass
of Calcium Phosphate could be formed?
Plan: (1) Calculate moles of P4.
(2) Use molar ratios to get moles of Ca3(PO4)2.
(3) Convert the moles of product back into mass by using the
molar mass of Calcium Phosphate.
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - II
Solution:
moles of P4 =
For Reaction #1 [ 4 P4 + 10 KClO4
For Reaction #2 [ 1 P4O10 + 6 H2O
For Reaction #3 [ 2 H3PO4 + 3 Ca(OH)2
moles Ca3(PO4)2 =
4 P4O10 + 10 KCl ]
4 H3PO4 ]
1 Ca3(PO4)2 + 6 H2O]
Calculating the Amounts of Reactants and
Products in a Reaction Sequence - III
Molar mass of Ca3(PO4)2 = 310.18 g mole
mass of product =
Balanced
reaction!
Defines
stoichiometric
ratios!
Unbalanced (i.e., non-stoichiometric)
mixture!
Limited by syrup!
Limiting Reactant
In a chemical reaction where arbitrary amounts
of reactants are mixed and allowed to react, the
one that is used up first is the limiting reactant.
A portion of the other reactants remains.
There is a systematic procedure for finding the
limiting reagent based on the reactant ratio (RR)
defined as the ratio of the number of moles of a
reactant to its coefficient in a balanced chemical
equation. The reagent with the smallest reactant
ratio is the limiting reactant.
For a Reaction of the Form:
aA + bB = cC + dD
If compounds A and B are present in the mole
amounts called for in the balanced reaction, then the
following equation is valid:
moles of A a

moles of B b
or
moles of A moles of B

 RR
a
b
If the Reactant Ratios of all reactants
are equal, then there is no limiting
reactant, and all reactants will be
consumed.
If all reactant ratios are not equal,
then the reactant with the smallest
reactant ratio is the limiting reactant.
To proceed, first calculate the reactant ratios for all of the
reactants:
starting moles A
RR A 
,
a
starting moles B
RR B 
, ...
b
From among these, choose (RR)min, the smallest reactant
ratio. This identifies the limiting reactant.
Once (RR)min is identified, it can be used to calculate the
actual moles of other reactants and products actually
consumed, e.g.
Actual moles of B consumed = (RR)minb
Actual moles of C produced = (RR)minc
Problem 9-3: Acid - Metal Limiting Reactant - I
• 2Al(s) + 6HCl(g)
2AlCl3(s) + 3H2(g)
Consider the reaction above. If we react 30.0 g Al and
20.0 g HCl, how many moles of aluminum chloride
will be formed?
• 30.0 g Al
• 20.0g HCl
• Limiting reactant = reactant with RRmin =
Problem 9-3: Acid - Metal Limiting Reactant - II
Calculate the yield of AlCl3 from:
Moles AlCl3 = (RR)min x reaction coefficient
of AlCl3.
Problem 9-4: Ostwald Process Limiting Reactant
What mass of NO could be formed by the reaction 30.0g of
ammonia gas and 40.0g of oxygen gas w/ the rxn below?
4NH3 (g) + 5 O2 (g)
4NO(g) + 6 H2O(g)
30.0g NH3
40.0g O2
RRmin =_______, therefore ______ is the Limiting Reactant!
Moles NO formed =
Mass NO =
Chemical Reactions in Practice: Theoretical,
Actual, and Percent Yields
Theoretical yield: The amount of product indicated by the
stoichiometrically equivalent molar ratio in the balanced equation.
Side Reactions: These form different products that
take away from the theoretical yield of the main product.
Actual yield: The amount of product that is actually obtained.
Percent yield (%Yield):
% Yield =
or
Actual Yield (mass or moles)
x 100%
Theoretical Yield (mass or moles)
Actual yield = Theoretical yield x (% Yield / 100%)
Problem 9-5: Percent Yield
Problem: The chemical reaction between iron and water to form
the iron oxide, Fe3O4 and hydrogen gas is given below. If 4.55 g of iron is
reacted with sufficient water to react all of the iron to form rust, what is
the percent yield if only 6.02 g of the oxide are formed?
Plan: Calculate the theoretical yield and use it to calculate the percent
yield, using the actual yield.
Solution:
3 Fe(s) + 4 H2O(l)
Fe3O4 (s) + 4 H2 (g)
Moles Fe =
Theoretical moles Fe3O4 =
Theoretical mass Fe3O4 =
Percent Yield = Actual Yield x 100% =
Theoretical Yield
Problem 9-6: Percent Yield / Limiting Reactant - I
Problem: Ammonia is produced by the Haber process using nitrogen
and hydrogen Gas. If 85.90g of nitrogen are reacted with
21.66 g hydrogen and the reaction yielded 98.67 g of
ammonia what was the percent yield of the reaction.
N2 (g) + 3 H2 (g)
2 NH3 (g)
Plan: Since we are given the masses of both reactants, this is a limiting
reactant problem. First determine which is the limiting reactant
then calculate the theoretical yield, and then the percent yield.
Solution:
moles N2 =
moles H2 =
Problem 9-6: Percent Yield / Limiting Reactant - II
Solution Cont.
N2 (g) + 3 H2 (g)
2 NH3 (g)
We have 3.066 moles of Nitrogen, and it is limiting, therefore the
theoretical yield of ammonia is:
mol NH3 =
mass NH3 =
Percent Yield =
Percent Yield =
Actual Yield x 100%
Theoretical Yield
98.67 g NH3
g NH3
x 100% =
Answers to Problems in Lecture # 9
1. (a) 2.629 moles H2O; (b) 44.81g H2S, 68.36g Al(OH)3
2. 77.61 g Ca3(PO4)2
3. 0.183 mol of AlCl3
4. 30.0 g NO
5. 95.6 %
6. 94.49 %