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Limiting Reagents
Caution: this stuff is difficult to follow at first.
Be patient.
g of
NaHCO3
mL of 3M
HCl
1 10 252 10 503 10 100
How can we prove that our conclusions about limiting reagents is correct?
Balloon & Flask Demonstration
Limiting reagent definedLimiting reagent defined
Q - How many moles of NO are produced if __ mol NH3 are burned in __ mol O2?
4 mol NH3, 5 mol O2
4 mol NH3, 20 mol O2
8 mol NH3, 20 mol O2
Given: 4NH3 + 5O2 6H2O + 4NO
4 mol NO, works out exactly 4 mol NO, with leftover O2
8 mol NO, with leftover O2
• Here, NH3 limits the production of NO; if there was more NH3, more NO would be produced
• Thus, NH3 is called the “limiting reagent”4 mol NH3, 2.5 mol O2
• In limiting reagent questions we use the limiting reagent as the “given quantity” and ignore the reagent that is in excess …
2 mol NO, leftover NH3
Limiting reagents in stoichiometryLimiting reagents in stoichiometry
E.g. How many grams of NO are produced if 4 moles NH3 are burned in 20 mol O2?
Since NH3 is the limiting reagent we will use this as our “given quantity” in the calculation
4NH3 + 5O2 6H2O + 4NO
4 mol NO 4 mol NH3
x
# g NO=4 mol NH3 = 120 g NO
30.0 g NO1 mol NO
x
• Sometimes the question is more complicated. For example, if grams of the two reactants are given instead of moles we must first determine moles, then decide which is limiting …
Solving Limiting reagents 1: g to molSolving Limiting reagents 1: g to mol
Q - How many g NO are produced if 20 g NH3 is burned in 30 g O2?
A - First we need to calculate the number of moles of each reactant
4NH3 + 5O2 6H2O + 4NO
1 mol NH3 17.0 g NH3
x # mol NH3= 20 g NH3 1.176 mol NH3
=
1 mol O2 32.0 g O2
x # mol O2= 30 g O20.9375 mol O2
=
A – Once the number of moles of each is calculated we can determine the limiting reagent via a chart …
NH3 O2
What we have
What we need
1.176 0.937 1.176/0.937 =
1.25 mol 0.937/0.937
= 1 mol
*Choose the smallest value to divide each by** You should have “1 mol” in the same column
twice in order to make a comparison
4 5
4/5 = 0.8 mol 5/5 = 1 mol
A - There is more NH3 (what we have) than needed (what we need). Thus NH3 is in excess, and O2 is the limiting reagent.
2: Comparison chart2: Comparison chart
3: Stoichiometry (given = limiting)3: Stoichiometry (given = limiting)So far we have followed two steps … 1) Expressed all chemical quantities as moles2) Determined the limiting reagent via a chart Finally we need to …3) Perform the stoichiometry using the limiting
reagent as the “given” quantity Q - How many g NO are produced if 20 g NH3
is burned in 30 g O2? 4NH3 + 5O2 6H2O + 4NO
4 mol NO5 mol O2
x
# g NO=30 g O2
22.5 g NO=
30.0 g NO1 mol NO
x 1 mol O2 32.0 g O2
x
Limiting Reagents: shortcutLimiting Reagents: shortcut• Limiting reagent problems can be solved
another way (without using a chart)…• Do two separate calculations using both given
quantities. The smaller answer is correct.Q - How many g NO are produced if 20 g NH3 is
burned in 30 g O2? 4NH3 + 5O2 6H2O+ 4NO
4 mol NO5 mol O2
x 30 g O2
22.5 g NO=
30.0 g NO1 mol NO
x 1 mol O2 32.0 g O2
x
4 mol NO4 mol NH3
x
# g NO=20 g NH3
35.3 g NO=
30.0 g NO1 mol NO
x 1 mol NH3 17.0 g NH3
x
Practice questionsPractice questions1. 2Al + 6HCl 2AlCl3 + 3H2
If 25 g of aluminum was added to 90 g of HCl, what mass of H2 will be produced (try this two ways – with a chart & using the shortcut)?
2. N2 + 3H2 2NH3: If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?
3. What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?
4. When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g of O2, how much CO2 is produced?
5. How can you tell if a question is a limiting reagent question vs. typical stoichiometry?
11 1 mol Al 27.0 g Al
x # mol Al =25 g Al = 0.926 mol
# mol HCl = 90 g HCl 1 mol HCl 36.5 g HCl
x = 2.466 mol
Al HCl
What we
have
What we
need
0.926 2.466 0.926/0.926
= 1 mol 2.466/0.926
= 2.7 mol
2 6 2/2 = 1 mol 6/2 = 3 mol
HCl is limiting.
3 mol H2
6 mol HClx
# g H2 =90 g HCl 2.0 g H2
1 mol H2
x 1 mol HCl 36.5 g HCl
x = 2.47 g H2
Question 1: shortcutQuestion 1: shortcut
2Al + 6HCl 2AlCl3 + 3H2
If 25 g aluminum was added to 90 g HCl, what mass
of H2 will be produced?
3 mol H2
2 mol Al x # g H2= 25 g Al = 2.78 g H2
2.0 g H2
1 mol H2
x 1 mol Al27.0 g Al
x
3 mol H2
6 mol HClx # g H2 = 90 g HCl = 2.47 g H2
2.0 g H2
1 mol H2
x 1 mol HCl36.5 g HCl
x
Question 2: shortcutQuestion 2: shortcut
N2 + 3H2 2NH3
If you have 20 g of N2 and 5.0 g of H2, which is the limiting reagent?
2 mol NH3
1 mol N2 x
# g NH3=
20 g N2 = 24.3 g H2 17.0 g NH3
1 mol NH3
x 1 mol N2
28.0 g N2
x
2 mol NH3
3 mol H2
x
# g NH3 =5.0 g H2 = 28.3 g H2
17.0 g NH3
1 mol NH3
x 1 mol H2
2.0 g H2
x
N2 is the limiting reagent
Question 3: shortcutQuestion 3: shortcut
4Al + 3O2 2 Al2O3
What mass of aluminum oxide is formed when 10.0 g of Al is burned in 20.0 g of O2?
2 mol Al2O3
4 mol Al x
# g Al2O3=
10.0 g Al = 18.9 g Al2O3 102.0 g Al2O3
1 mol H2
x 1 mol Al27.0 g Al
x
2 mol Al2O3
3 mol O2
x
# g Al2O3=
20.0 g O2 = 42.5 g Al2O3 102.0 g Al2O3
1 mol H2
x 1 mol O2
32.0 g O2
x
Question 4: shortcutQuestion 4: shortcutC3H8 + 5O2 3CO2 + 4H2O
When C3H8 burns in oxygen, CO2 and H2O are produced. If 15.0 g of C3H8 reacts with 60.0 g
of O2, how much CO2 is produced?3 mol CO2
1 mol C3H8 x
# g CO2=15.0 g C3H8 = 45.0 g CO2
44.0 g CO2
1 mol CO2
x 1 mol C3H8
44.0 g C3H8
x
3 mol CO2
5 mol O2
x
# g CO2=60.0 g O2 = 49.5 g CO2
44.0 g CO2
1 mol CO2
x 1 mol O2
32.0 g O2
x
5. Limiting reagent questions give values for two or more reagents (not just one)
N2 H2
What we have
What we need
Question 2Question 2
0.714 mol 2.5 mol
0.714/0.714 = 1 mol
2.5/0.714 = 3.5 mol
We have more H2 than what we need, thus H2 is in excess and N2 is the limiting factor.
1 mol 3 mol
1 mol N2 28 g N2
x # mol N2= 20 g N2 0.714 mol N2=
1 mol H2 2 g H2
x # mol H2= 5.0 g H2 2.5 mol H2=
Al O2
33 4Al + 3O2 2 Al2O3
1 mol Al 27 g Al
x # mol Al = 10 g Al 0.37 mol Al=
1 mol O2 32 g O2
x # mol O2 = 20 g O20.625 mol O2=
0.37 mol 0.625 mol 0.37/.37 =
1 mol 0.625/0.37
= 1.68 mol
4 mol 3 mol 4/4 = 1 mol 3/4 = 0.75 mol
What we have
What we need
There is more than enough O2; Al is limiting
2 mol Al2O3
4 mol Al x # g Al2O3 = 0.37 mol Al
18.87 g Al2O3=
102 g Al2O3
1 mol Al2O3
x
C3H8 O2
44 C3H8 + 5O2 3CO2 + 4H2O1 mol C3H8 44 g C3H8
x # mol C3H8 = 15 g C3H80.34 mol
C3H8
=
1 mol O2 32 g O2
x # mol O2 = 60 g O21.875 mol O2=
0.34 mol 1.875 mol 0.34/.34 = 1
mol 1.875/0.34
= 5.5 mol
1 mol 5 mol
What we have
Need
3 mol CO2
1 mol C3H8 x
# g CO2 =0.34 mol C3H8
45 g CO2=
44 g CO2
1 mol CO2
x
We have more than enough O2, C3H8 is limiting
Limiting Reagents: shortcutLimiting Reagents: shortcut
MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgClIf 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced?
2 mol AgCl1 mol MgCl2
x
# g AgCl=25 g MgCl2
75.25 g AgCl=
143.3 g AgCl1 mol AgCl
x 1 mol MgCl295.21 g MgCl2
x
2 mol AgCl2 mol AgNO3
x
# g AgCl=68 g AgNO3
57.36 g AgCl=
143.3 g AgCl1 mol AgCl
x 1 mol AgNO3
169.88 g AgNO3
x
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