STOICHIOMETRY (b) how many moles of potassium sulfate
are produced from 8 moles of
Stoichiometry potassium nitrate?
deals with the combining proportions of the (c) how many moles of nitric acid will be
elements or compounds involved in chemical produced from 50 moles of sulfuric
reactions acid?
Mass-Mass Relationship
tells us the relationship of the moles & masses
of the reactant to one another & also to the recall that:
products moles = mass/molecular mass
In the combustion of triethylene glycol, sample problem:
2C6H14O4(l) + 15O2(g) 12CO2(g) + 14H2O(l) based on the balanced chemical equation:
for the reaction to undergo, 2 moles of 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
C6H14O4(l) must react with 15 moles of O2(g) to how many grams of NH3 will be required to
produce 12 moles of CO2(g) and 14 moles of produce 150 g of H2O?
H2O(l). g NH3 = 150 g H2O1 mol H2O 4 mol NH3
18 g H2O 6 mol H2O
Mole-Mole Relationship 17 g NH3
1 mol NH3
sample problem:
= 94.44 g NH3
Consider the reaction,
2C6H14O4(l) + 15O2(g) 12CO2(g) + 14H2O(l) Exercises:
(a) How many moles of CO2 would be 1. What is the mass in grams of H3PO4 produced
produced if there are 20 moles of from the reaction of 500 g of P4O10 with H2O?
C6H14O4? 12 mol CO2
P4O10 + 6H2O 4H3PO4
mol CO2 = 20 mol C6H14O24 mol
2. The reaction between hydrogen sulfite &
C6H14 O4 calcium hydroxide produced calcium sulfite &
water.
= 120 mol CO2
(a) formulate the balanced chemical
(b) How many moles of O2 are needed to equation
produce 75 moles of H2O? (b) how many moles of calcium sulfite will
mol O2 = 75 mol H2O 15 mol O be produced when 150 g of hydrogen
2
14 mol H2O sulfite was reacted to calcium
hydroxide?
= 80.36 mol O2 (c) what is the mass of each of the
(c) How many moles of H2O are produced if products produced in the reaction?
there are 100 moles of CO2 produced?
mol H2O = 100 mol CO214 mol H2O Percent Purity
12 mol CO2 various reactants in chemical reactions are not
pure (i.e., mixed with other useless contents)
= 116.67 mol H2O & only pure or reactive compounds participate
Exercises: in reactions
1. From the balanced equation,
2Al(s) + 6HCl(aq) 2AlCl3(aq) + 3H2(g)
the purity of a substance is expressed as
percent purity, given as:
(a) How many moles of Al are necessary to
react with 3.50 moles of HCl?
% purity = mass of pure compound in a substance
(b) How many moles each of AlCl3 & H2 are x 100
produced? total mass of the substance
2. The reaction is given as: potassium nitrate +
sulfuric acid potassium sulfate + nitric acid total mass of the substance = mass of the pure
(a) formulate the balanced equation compound + mass of impurities
Gen Chem 113
Mike Filomeno
Stoichiometry
Page 1 of 3
sample problem:
how many grams of sodium sulfate could be
produced from 750 kilograms of 85% pure = 693.75 kg C8H4O3
NaCl?
2NaCl + H2SO4 Na2SO4 + 2HCl % yield = actual yield x 100
mass of NaCl = 0.85 x (500 kg) theoretical yield
= 425 kg NaCl or 425,000 g NaCl
g Na2SO4 = 425 kg NaCl1000 g actual yield = (% yield x theoretical yield) / 100
1 mol NaCl = (65%)(693.75)/100
1 kg 58.5 g NaCl = 450.94 kg C8H4O3
Exercises:
119 g
1 mol Na2SO4 1. Nitroglycerin (C3H5N3O9) is a powerful
Na2SO4 explosive. Its decomposition may be
1 mol represented by:
2 mol NaCl Na2SO4
4 C3H5N3O9 6N2 + 12CO2 + 10H2O + O2
= 432264.96 g or 432.26 kg (a) what is the maximum amount of O2 in
grams that can be obtained from 2.00 x
102 g of nitroglycerin?
Exercises: (b) calculate the percent yield in this
1. A 300 sample of impure Zinc reacts with 125 reaction if the amount of O2 generated
mL of 25% HCl solution (density of acid is 1.18 is found to be 6.55 g.
g/mL). Calculate the % Zinc in the sample.
2. Titanium is prepared by the reaction of
2. For the reaction: CaO + 3C CaC2 + CO. The titanium(IV) chloride with molten magnesium
crude product from this reaction is typically between 950°950°C & 1150°1150°C according to the
65% CaC2. How much CaO must be reacted to reaction:
produce for 50 kg of CaC2?
TiCl4(g) + 2Mg(l) Ti(s) + 2MgCl2(l)
If 200 kg of Mg is reacted with TiCl4:
Theoretical & Actual Yield (a) calculate the theoretical yield of Ti(s)
during chemical reactions, not all reactants are (b) determine the actual yield of Ti(s) if the
percent yield in this reaction 68%
totally converted to products. Some of them
are left unreacted since not all reactions
proceed to completion.
Limiting & Excess Reactants
Theoretical Yield is the amount of product obtained in chemical reactions, reactants are not
if the reaction goes to completion. (i.e., all reactants present in exact stoichiometric amounts.
are converted into products) It is the maximum Some of the reactants will be used up
amount which can be produced during a reaction. completely while others will be left over at the
end of a reaction.
Actual Yield is the amount of product that is actually
obtained during a reaction. (actual yield is always Limiting Reactant is the reactant that reacts
less than the theoretical yield) completely (or used up first) in a chemical reaction.
The extent of a reaction depends highly on the
% yield = actual yield x 100 amount of this reactant.
theoretical yield
Excess Reactant is the reactant present in
sample problem: quantities greater than necessary to react with the
Phthalic anhydride, C8H4O3, is made by the controlled quantity of the limiting reactant.
oxidation of naphthalene, C10H8, by the reaction:
2C10H8 + 9O2 2 C8H4O3 +4CO2 + 4 H2O Guidelines in Determining the Limiting
Since some of the naphthalene is oxidized to other Reactant:
products, only 65% of the maximum yield was 1. Select a reactant which will be used as the
obtained. How much phthalic anhydride would be basis for calculation
produced from the oxidation of 600 kg of C10H8? 2. Using the basis, calculate the theoretical mass
of the other reactant
theoretical yield of C8H4O3 3. Compare the theoretical mass to the given
= 600 kg C10H8 1 kmol C10 H8 2 kmol C8H4O3 actual mass of the reactant.
128 kg C10 H8 2 kmol C10 H8
Gen Chem 113
Mike Filomeno 148 kg C8H4O3
Stoichiometry
1 kmol C8H4O3 Page 2 of 3
4. If the theoretical mass is greater than the 1. Urea [(NH2)2CO] is prepared by reacting
actual mass, that reactant is already the ammonia with carbon dioxide:
limiting reactant. 2NH3(g) + CO2(g) (NH2)2CO(aq) + H2O(l)
5. If the theoretical mass is less than the actual In one process, 630 g of ammonia are allowed
mass, that reactant is the excess reactant. to react with 1140 g of CO2.
(a) determine the limiting & excess reactants
Guidelines in Determining the Amount of Excess (b) calculate the mass of urea formed
Reactant Left after Reaction: (c) how much of the excess reactant (in
1. Determine the actual amount of the excess grams) is left after the reaction?
reactant that will react completely with the 2. The reaction between aluminum & iron(III)
limiting reactant oxide can generate temperatures approaching
2. Subtract this amount from the actual amount 3000°C and is used in welding metals:
3000°
given 2Al + Fe2O3 Al2O3 + 2Fe
If 124 g of Al are allowed to react with 600 g of
sample problem: Fe2O3:
Flourine reacts with iron to produce iron(III) fluoride. (a) determine the limiting & excess reactants
If 5.0 g of fluorine is added to 10 g of iron: (b) calculate the mass of Al2O3 formed
(a) determine the limiting & excess reactants (c) how much of the excess reactant (in
(b) calculate the mass of iron(III) fluoride obtained grams) is left at the end of the reaction?
from the reaction 3. Industrially, vanadium metal, which is used in
(c) how much of the excess reagent will be left steel alloys, can be obtained by reacting
unreacted? vanadium(V) oxide with calcium at high
temperatures:
basis: Iron (Fe) 5Ca + V2O5 5CaO + 2V
2Fe + 3F2 2FeF3 In one process, 1.54 x 103 g of V2O5 react with
1.96 x 103 g of Ca.
mass of F2 = 10 g Femol Fe (a) calculate the theoretical yield of V
3 mol F2
56 g Fe (b) calculate the percent yield if 803 g of V are
2 mol Fe obtained
38 g F2
1 mol F2
= 10.18 g F2 (theoretical)
since the theoretical (10.18 g) > actual (5.0 g),
F2 is the limiting reactant
it follows that Fe is the excess reactant
mass of FeF3 = 5.0 g F2 1 mol F2 2 mol FeF3
38 g F2
3 mol F2
113 g FeF3
1 mol FeF3
= 9.91 g FeF3
mass of Fe reacted = 5.0 g F21 mol F2
2 mol Fe
38 g F2 3 mol F2
56 g Fe
1 mol Fe
= 4.91 g Fe
excess Fe = (10.0 g – 4.91 g)
= 5.10 g
Exercises:
Gen Chem 113
Mike Filomeno
Stoichiometry
Page 3 of 3
