Transcript Information Given by Chemical Equations

Chemical Equation
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Describes a chemical reaction
A+BC
A and B = reactants
C = product
CO(g) + 2H2(g)  CH3OH(l)
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Information Given by
Chemical Equations
CO(g) + 2H2(g)  CH3OH(l)
• 1 molecule of CO + 2 molecules of
H2  one molecule of CH3OH
• 10 molecules of CO + 20 molecules
of H2  10 molecules of CH3OH
• 1 mole of CO + 2 moles of H2  1
mole of CH3OH
• Do you notice a relationship pattern?
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Stoichiometry
• Process of using a
chemical equation to
calculate the relative
masses of reactants and
products involved in a
reaction.
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Balancing equations
• In a chemical reaction, atoms are
neither created nor destroyed.
• All atoms present in the reactants
must be accounted for among the
products.
• Therefore, one needs to balance
the chemical equation for the
reaction.
• What goes in must come out,
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albeit in a different form.
Rules for balancing
equations
1. Balance out metals and polyatomic ions first
• Consider polyatomic ions as a single species
2. Balance out elements that are found in more than
one species last
• Leave H for last: Na + H2O  NaOH + H2
3. Don’t get discouraged!
• Balancing one atom can throw others off
• Balance the equation afresh
4. Do NOT change the nature of the compound
• To balance H2O you cannot make it H2O2
• Why not?
5. Thus, put numbers in front of compounds
6. Lowest whole numbers
• Incorrect: 4A + 4B  4AB
• Correct: A + B  AB
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Example
• Balance:
• Na(s) + H2O(l)  NaOH(aq) + H2(g)
• Na is balanced, O is balanced, but H is not
•  balance H  put 2 in front of NaOH
• This gives us 4 H on the products side
• Add 2 in front of water on reactants side
• H’s are now balanced
− So are O’s (2 on each side)
• Na is now unbalanced
• Add a 2 to Na on reactant side
• Recapitulate: check equation
• 2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
• BALANCED!
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Balance the following
• Al(s) + F2(g)  AlF3(s)
• KClO3(s)  KCl(s) + O2(g)
• KI(aq) + Pb(NO3)2(aq)  KNO3(aq) + PbI2(s)
• C2H6(g) + O2(g)  CO2(g) + H2O(l)
• Let’s talk about a neat trick for the
above
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Mole-Mole
Relationships:
Dimensional Analysis
Consider the decomposition of water:
2H2O(l)  2H2(g) + O2(g)
• If we decompose 4 mol of H2O, how many
moles of O2 do we get?
• If we decompose 5.8 mol of H2O, how
many moles of H2 do we get?
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More practice
• Calculate the moles of C3H8 used when
4.30 moles of CO2 are obtained.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
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Mass Calculations
Steps for Calculating the Masses of Reactants and
Products in Chemical Reactions
1. Balance the equation for the reaction.
2. Convert the masses of reactants or
products to moles.
3. Use the balanced equation to set up the
appropriate mole ratio(s).
4. Use the mole ratio(s) to calculate the
number of moles of the desired reactant
or product.
5. Convert from moles to mass.
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In other words…
• gramsA  molesA
• molesA  molesB
• molesB  gramsB
• Do it all as a dimensional
analysis problem!
• See next slide
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Example
What mass of 2 should
we weigh out to react
with 35.0 g Al?
2Al(s) + 32(s)  2Al3(s)
mol
) = 1.30 mol Al
26.98154 g
3 mol I2
1.30 mol Al  (
) = 1.95 mol I 2
2 mol Al
253.8090 g
1.95 mol I2  (
) = 495 g I 2
mol
35.0 g Al  (
• gramsA  molesA:
• molesA  molesB:
• molesB  gramsB:
• In one-step:
35.0 g Al  (
3 mol I 2
mol
253.8090 g
) (
) (
) = 495 g I2
26.98154 g
2 mol Al
mol
• Avoids rounding errors
too!
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Practice
• What mass of CO2 is produced when 96.1 g
of C3H8 react with sufficient O2.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
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Calculations Involving
a Limiting Reactant
• When there is insufficient amount of one of
the reactants.
• Calculate the mass required for the
product in question using both reactants.
• Which ever gives the least amount of
product is the limiting reactant.
• The mass of product obtained from using
the limiting reactant is the correct
amount.
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Example
What mass of AlI3 will we obtain if we react
35.0 g Al with 400.0 g I2?
2Al(s) + 3I2(s)  2AlI3(s)
• 35.0 g Al x (mol/26.98154 g) x (2 mol AlI3/2
mol Al) x (407.6950 g/mol) = 529 g AlI3
• Versus: 400.0 g I2 x (mol/253.8090 g I2) x (2
mol AlI3/3 mol I2) x (407.6950 g/mol) = 428
g AlI3
•  I2 is the limiting reactant.
• The mass of AlI3 obtained is 428 g.
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Practice
Example
Calculate the mass of lithium nitride formed
from 56.0 g of N2 and 56.0 g of Li in the
balanced equation:
6Li(s) + N2(g)  2Li3N(s)
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More practice
Example
Calculate the mass of aluminum sulfate
formed from 50.0 g of Al and 50.0 g of
H2SO4 in the balanced equation:
2Al(s) + 3H2SO4(aq)  Al2(SO4)3(s) + 3H2(g)
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Percent Yield
• Theoretical Yield: Amount of product
predicted from the amounts of
reactants used.
• Actual Yield: Amount of product
actually obtained. Actual yields are
determined by doing the reaction
• __Actual Yield__ X 100% = Percent Yield
Theoretical Yield
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Percent Yield
Consider the reaction:
TiCl4(g) + O2(g)  TiO2(s) + 2Cl2(g)
(a) Suppose 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.
(b) If 2.12 x 103 g was produced in the
laboratory, what is the percent yield?
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