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Stoichiometry Advanced
HW: Read CH 6 and
CH 9
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Classifying Reactions
 Synthesis
(combination): 2 or more reactants
come together to form one product
 Decomposition: 1
compound breaks down into 2
or more smaller compounds
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Classifying Reactions

Single displacement (replacement): An uncombined
element switches with another element in a
compound

Double displacement (replacement): 2 compounds
“switch partners” with each other

***Combustion: involves oxygen gas as one of the
reactants (it usually produces carbon dioxide and
water)
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YOU TRY!

Balance the following equations and indicate the type of
reaction taking place

A)
NaBr + H3PO4  Na3PO4 + HBr

B)
H2CO3  H+ +CO32-

C)
As + Cl2  AsCl3
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Limiting Reactants
 The
reactant that runs out first and thus limits the
amounts of products that can form is called the limiting
reactant (or limiting reagent)
+ Limiting Reactant problem-solving
plan

(For limiting reactant problems, the problem will give you information
about two reactants)

A. Start with a balanced equation.

B. First determine which reactant will run out (which reactant is the limiting
reactant)


1) You must change all starting masses into moles.
2) The trick to determine which reactant limits is to divide the moles of each reactant by
the coefficient (from the balanced equation) associated with that reactant. The number
that comes out the smallest indicates which reactant is the limiting one. The limiting
reactant is the one that you must base all your other calculations on because it is the
substance that limits how much of everything else can be made or is needed.

C. The other reactant (if there’s only two) will be the excess reactant,
and some of it will be left over.

**Use molar ratios to convert the moles of the known limiting substance
to the moles of other compounds (use coefficients from balanced
equation) to answer more questions.
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Limiting Reactants
 Fe
+ CuSO4 
FeSO4 + Cu

If 120. g of Fe are reacted with 200. g of CuSO4, identify the limiting
reagent. Which reagent is in excess?

120 g Fe x 1 mol Fe = 2.15 mol Fe / 1
= 2.15 (excess)
55.85 g Fe

200 g CuSO4 x 1 mol CuSO4 = 1.25 mol CuSO4 /1 = 1.25 (limiting)
159.62 g CuSO4
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Limiting Reactant
 If
you are asked to determine the amount of excess
reactant leftover:
 A. Knowing
which reactant limits and which is
excess, use the limiting reactant to determine the
moles of the excess reactant that is actually
needed to do the reaction. Subtract.
 B. Convert
these moles into grams.
+ Limiting Reactants

Fe + CuSO4 
FeSO4 + Cu

If 120. g of Fe are reacted with 200. g of CuSO4, identify the limiting
reagent. Which reagent is in excess? Calculate the mass of Copper
formed. How much of the excess reagent is left over at the end of the
reaction?

120 g Fe x 1 mol Fe
= 2.15 mol Fe (excess)
55.85 g Fe

200 g CuSO4 x
1 mol CuSO4
= 1.25 mol CuSO4 (limiting)
159.62 g CuSO4

1.25 mol CuSO4 x 1 mol Cu
x
1 mol CuSO4

63.55 g Cu = 79.4 g Cu
1 mol Cu
Out of 2.15 mol Fe, only 1.25 mol is used so 0.9 mol Fe in excess

0.9 mol Fe 58.85 g Fe = 53.0 grams in excess
1 mol Fe
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Limiting Reactants

YOU TRY! (Ouch! There is LOTS to remember)
 Which
is the limiting reactant? Calculate the mass of
lithium nitride formed from 56.0 g of N2 gas and 56.0 g
of Li in the unbalanced reaction:
Li +
N2 
Li3N
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Limiting Reactants

YOU TRY! (I think we should try another)
 Methanol, CH3OH, which
is used as a fuel in racing cars
can be made by the reaction of carbon monoxide and
hydrogen gas. Suppose 356 g of CO and 65.0 g of H2
are mixed and allowed to react. A) What mass of
methanol can be produced? B) What mass of the excess
reactant remains after the limiting reactant has been
consumed?
CO(g) +
H2(g) 
CH3OH(l)
+ Yield

The maximum amount of product that can be produced
is called the theoretical yield.

You can compare the calculated yield with the
actual/experimental yield and find the percent yield.
(how far off from the calculated value was your
experiment)
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Percent Yield
 Percent
yield = actual yield x 100%
theoretical yield
In previous you try example, 93.7 g Li3N should have
formed. If this reaction actually gave 55.2 g Li3N,
what is the percent yield of Li3N?
55.2 g Li3N x 100 = 58.9%
93.7 g Li3N
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Percent Yield
 YOU TRY!
Good Luck!
TiCl4 + O2  TiO2 + 2Cl2
Suppose that 6.71 x 103 g of TiCl4 is reacted with 2.45 x
103 g of O2. Calculate the maximum mass of TiO2 that can
form. If the percent yield of TiO2 is 75%, what mass is
actually formed?