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PRODUCTS
WHY CAN’T I USE ALL MY BUILDING
BLOCKS? (REACTANTS)
•WE DON’T HAVE ENOUGH OF EACH OF THE BLOCKS TO BUILD FULL PRODUCTS
•SOME REACTANTS ARE LEFT OVER BUT ONE REACTANT GETS COMPLETELY USED UP!
Limiting Reagents
STOICHIOMETRY
•the the reactantreactant which is totally which is totally consumed when the chemical consumed when the chemical reaction is completereaction is complete
Limiting Reagent
“the amount of product formed is limited by this reactant”
Limiting reagentLimiting reagent
Q - How many moles of NO are produced if 4 mol NH3 are burned in 5 mol O2?
Given: 4NH3 + 5O2 6H2O + 4NO
4 mol NO, works out PERFECTLY – both reactants are completely used up
Here, NH3 limits the production of NO; If there was more NH3, more NO would be produced
NH3 is called the “limiting reagent” and O2 is in “excess
4 mol NO, with leftover O2
Given: 4NH3 + 5O2 6H2O + 4NO
Q - How many moles of NO are produced if 4 mol NH3 are burned in 20 mol O2?
On your worksheet: How many moles of NO are produced if
4 mol NH3 are burned in 2.5 mol O2?
2 mol NO, with leftover NH3
• Here, O2 limits the production of NO; if there was more O2, more NO would be produced
• Thus, O2 is called the “limiting reagent” and NH3 is in excess!
How can I tell which reactant is the limiting
reagent?Use a comparison chart between what
we have and what the balanced equation says we need…
NH3 O2
What we have Mole given Mole given
From the question
Mole ratio calculated
What we need Ratio in balanced equation
Ok… let’s try it!!4NH3 + 5O2 6H2O + 4NO
3.2 mol NH3 reacts with 1.6 mol O2
-which reactant will limit the production of the reactants?
NH3 O2
What we have
What we need
Comparison chartComparison chart
3.2 1.6
3.2/1.6 = 2 mol
1.6/1.6 = 1 mol
4/5 = 0.8 5/5 = 1
There is more NH3 than needed to react all the O2.
So O2 is the limiting reagent which makes NH3 the
excess reagent!• Now you can use the limiting
reagent moles to calculate how much product you can make!
Limiting reagents in stoichiometryLimiting reagents in stoichiometry
How many moles of NO are produced if 0.25 moles NH3 are burned in 0.56 mol O2? (make a chart)
4NH3 + 5O2 6H2O + 4NO
4 mol NO 4 mol NH3
x
NO
mol= 0.25 mol NH3 =0.25 mol NO
0.25 mol NH3 is the limiting reagent
Complete questions 3 and 4 on the worksheet!
Al(s) + MnO2(aq) Al2O3(aq) + Mn(s)
3. What is the limiting reactant when 0.1372 mol of aluminum reacts with 0.1264 mol of MnO2? How many moles of the aluminum product should be yielded from this reaction?
Al MnO2
What we have
What we need
0.1372 0.1264
0.1372/0.1264
1.085
0.1264/0.1264
1
3 4 2 3
4/3 = 1.33 3/3 = 1
So, based on the balanced equation, to use all the MnO2 I would need
more Aluminum than I have?
Aluminum is the LIMITING REACTANT!
Mol Al2O3 made = 0.1372 mol Al x 2 mol Al2O3
4 mol Al = 0.0686 mol
4. If 0.434 mol of both reactants are combined, which will be the limiting reagent?
What is the theoretical (predicted) maximum mass of manganese that can be yielded from this reaction?
LIMITING REACTANT IS aluminum
Theoretical maximum of Mn is 0.3255 mol which (using mm of Mn) converts to 17.88 g
•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 mass to moleSolving Limiting reagents mass to mole
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
=
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
4 5
Comparison chartComparison chart
A – Once the number of moles of each is calculated-find the LR…
A - There is more NH3 (what we have) than needed (what we need). Thus NH3 is in excess, and O2 is the limiting reagent.
StoichiometryStoichiometry1) Expressed all chemical quantities as moles
2) Determined the limiting reagent via a chart 3) Use the limiting reagent to determine how
much product can be made
Limiting Reagents: “shortcut”Limiting 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?
6. If 25 g magnesium chloride was added to 68 g silver nitrate, what mass of AgCl will be produced?
MgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl
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 25.0 g aluminum was added to 90.0 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
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=
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
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
18.9 g Al2O3=
102 g Al2O3
1 mol Al2O3
x
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
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
45.0 g CO2=
44 g CO2
1 mol CO2
x
We have more than enough O2, C3H8 is limiting
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)
Question 6: shortcutQuestion 6: shortcutMgCl2 + 2AgNO3 Mg(NO3)2 + 2AgCl
If 25.00 g magnesium chloride was added to 68.00 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
