pp. 326 – 334
 If
you are given one dozen loaves of bread, a
gallon of mustard, and three pieces of
salami, how many salami sandwiches can
you make?
 The limiting reagent is the reactant you run
out of first.
 The limiting reagent determines how much
product you can make
 The excess reagent is the one you have left
over.
 Nitrogen
and hydrogen molecules react
to form ammonia:
N2 + 3H2  2NH3
 Use
Fig. 7 p. 330 to answer the following:
a) Identify the limiting and excess reagents in the
flask.
b) What is the maximum number of ammonia
molecules that can be made?
c) How many molecules of excess reagent
remain?
 Step
1: Write a balanced chemical
equation listing given value(s) and
required value(s)
 Step 2: Determine the limiting reagent by
using the amount (n) of one reactant to
find the stoichiometric amount (n) of the
other.
 Step 3: Use the amount of limiting reagent
to find the amount of required substance.
 Determine
the amount of titanium metal
produced when 2.8 mol of titanium(IV)
chloride reacts with 5.4 mol of
magnesium.
 Step
G:
R:
1:
TiCl4 + 2Mg  Ti + 2MgCl2
2.8 mol
5.4 mol
nTi = ?
 Step
2:
nMg = 2.8 molTiCl4 × (2 molMg ÷ 1 molTiCl4)
nMg = 5.6 mol
 Since
5.6 mol is more than the 5.4 mol
given in the question, magnesium must
be limiting
 It doesn’t matter which of the two
reactants you choose to do this for, it will
give you the same limiting reagent.
 For
example:
nTiCl4 = 5.4 molMg × (1 molTiCl4 ÷ 2 molMg)
nTiCl4 = 2.7 mol
 Since
2.7 mol is less than the 2.8 mol
given in the question, magnesium must
be limiting
 If the calculated amout is less than the
given amout, it is execess; if it is more
than the given amout, it is limiting!
 Step
3:
nTi = 5.4 molMg × (1 molTi ÷ 2 molMg)
nTi = 2.7 mol
P: When 2.8 mol of titanium(IV)chloride is
combined with 5.4 mol of magnesium, 2.7
mol of titanium will be produced.
A
nitric acid spill is neutralized by adding
sodium hydrogen carbonate, NaHCO3(s):
HNO3(aq) + NaHCO3(s) → H2O(l) + CO2(g) + NaNO3(aq)
 What
amount of water is produced when 2.3
mol of nitric acid is combined with 2.0 mol
of sodium hydrogen carbonate?
[ans: 2.0 mol]
 This
follows the same strategy as for any
other stoichiometry problem involving
masses.
 The
only difference is that you first
determine which reactant is the limiting
reagent, then use the mass of the limiting
reagent to determine the masses of
product(s).
 What
mass of methanol can be produced
from 9.80 g of carbon monoxide & 1.30 g
of hydrogen?
 Step
1: Balanced Chemical Equation
CO(g)
+ 2H2(g)  CH3OH(l)
G: 9.80 g
1.30 g
28.01 g/mol 2.02 g/mol
R:
mCH3OH = ?
 Step
2: Convert mass of given substances to
amount (n) of given substances
nCO = 9.80 g ÷ 28.01 g/mol = 0.34988 mol
nH2 = 1.30 g ÷ 2.02 g/mol = 0.64356 mol
 Step
3: Determine the limiting reagent
nCO = 0.64356 molH2 × (1 molCO ÷ 2 molH2)
nCO = 0.32178 mol
Since the amount of CO present initially is
greater than this amount, CO is the excess
reagent.
 Step
4: Use the amount of the limiting
reagent to find the amount of required
substance.
nCH3OH = 0.64356 molH2 × (1 molCH3OH ÷ 2 molH2)
nCH3OH = 0.32178 mol
 Step
5: Convert amount (n) of required
substance to mass of required substance.
mCH3OH = 0.32178 mol × 32.05 g/mol
mCH3OH = 10.3 g
P: When 9.80 g of carbon monoxide reacts
with 1.30 g of hydrogen, 10.3 g of methanol
will be produced.
 Read
pp. 326 – 324
 Answer the following questions:
• p. 330 # 2, 3
• p. 335 # 1, 3a, 4, 6 – 10
 Read
investigation 7.5.1 “What Stopped
the Silver?” pp. 342 – 343
• Create a data table and be prepared to begin
this experiment tomorrow!