Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
3 H2(g)
First write a balanced
equation.
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Gram to Gram Conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Now let’s get organized.
Write the information
below the substances.
2
gram to gram conversions
Aluminum is an active metal that when placed in hydrochloric acid produces
hydrogen gas and aluminum chloride. How many grams of aluminum
chloride can be produced when 3.45 grams of aluminum are reacted with an
excess of hydrochloric acid?
2 Al(s) + 6HCl(aq)  2AlCl3(aq) +
? grams
3.45 g
3 H2(g)
Units match
3.45 g Al
mol Al
27.0 g Al
2 mol AlCl 3 133.3 g AlCl 3
mol AlCl 3
2 mol Al
Now
We must
Now
use
Let’s
the
always
use
work
molar
thethe
convert
molar
mass
problem.
ratio.
to
toconvert
moles.
to grams.
=
17.0 g AlCl3
3
4
Molarity
Molarity is a term used to express concentration. The units of molarity are
moles per liter (It is abbreviated as a capital M)
When working problems, it
is a good idea to change M
into its units.
moles
moles
M

Liter 1000 mL
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6
Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
What type of
problem(s) is
this?
Molarity
followed by
dilution.
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Solutions
A solution is prepared by dissolving 3.73 grams of AlCl3 in
water to form 200.0 mL solution. A 10.0 mL portion of the
solution is then used to prepare 100.0 mL of solution.
Determine the molarity of the final solution.
1st:
3.73 g
mol
= 0.140 mol
3
133.4 g 200.0 x 10 L
L
molar mass of AlCl3
2nd:
M1V1 = M2V2
dilution formula
(0.140 M)(10.0 mL) = (? M)(100.0 mL)
0.0140 M = M2 final concentration
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Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution are needed to
neutralize 35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
2 2O
____H
1 2SO4
+ ____Na
First write a balanced
Equation.
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Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
0.102 M mol
L
? mL
Our Goal
2 2O
____H
1 2SO4
+ ____Na
35.0 mL
0.125 mol 0.125 mol

L
1000 mL
Since 1 L = 1000 mL, we can use
this to save on the number of conversions
Now, let’s get organized. Place
numerical Information and
accompanying UNITS below each
compound.
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Solution Stoichiometry:
Determine how many mL of 0.102 M NaOH solution is needed to neutralize
35.0 mL of 0.125 M H2SO4 solution.
2
1 2SO4 
____NaOH
+ ____H
0.102 M mol
L
? mL
H2SO4
35.0 mL
2 2O
____H
1 2SO4
+ ____Na
35.0 mL
0.125 mol 0.125 mol

L
1000mL
H2SO4
0.125 mol
1000 mL
H2SO4
NaOH
2 mol
1 mol
H2SO4
Units Match
1000 mL NaOH = 85.8 mL NaOH
0.102 mol NaOH
Now let’s get to work
converting.
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13
Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
1st write out
a balanced chemical
equation
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Solution Stoichiometry
What volume of 0.40 M HCl solution is needed to
completely neutralize 47.1 mL of 0.75 M Ba(OH)2?
2HCl(aq) +
Ba(OH)2(aq) 
0.40 M
47.1 mL
0.75 M
? mL
Ba(OH)2
47.1 mL
0.75mol Ba(OH)2
2H2O(l) + BaCl2
Units match
HCl
2 mol
1000 mL Ba(OH)2 1 mol
Ba(OH)2
HCl
1000 mL
0.40 mol
HCl
= 176 mL HCl
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Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
2
1
2 2O(l) + ____BaCl
1
____HCl(aq)
+ ____Ba(OH)
2(aq)  ____H
2(aq)
25.00 mL
23.28 mL
0.135 mol
L
? mol
L
First write a balanced
chemical reaction.
17
Solution Stochiometry Problem:
A chemist performed a titration to standardize a barium hydroxide solution.
If it took 23.28 mL of 0.135 M hydrochloric acid to neutralize 25.00 mL of
the barium hydroxide solution, what was the concentration of the barium
hydroxide solution in moles per liter (M)?
2
1
2 2O(l) + ____BaCl
1
____HCl(aq)
+ ____Ba(OH)
2(aq)  ____H
2(aq)
25.00 mL
23.28 mL
Units match on top!
? mol
0.135 mol
L
L
23.28 mL HCl
25.00 x 10-3 L
Ba(OH)2
0.135 mol HCl l mol Ba(OH)2
= 0.0629 mol Ba(OH)2
1000 mL HCl 2 mol HCl
L Ba(OH)
2
Units Already Match on Bottom!
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Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
We must first
write a balanced
equation.
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Solution Stochiometry Problem:
48.0 mL of Ca(OH)2 solution was titrated with 19.2
mL of 0.385 M HNO3. Determine the molarity of
the Ca(OH)2 solution.
Ca(OH)2(aq) + 2 HNO3(aq)  2 H2O(l) + Ca(NO3)2(aq)
48.0 mL
?M
HNO3
19.2 mL
19.2 mL
0.385 M
0.385 mol

L
HNO 3
0.385 mol 1mol Ca(OH) 2
=0.0770 mol(Ca(OH)
1000 mL 2mol HNO 3 48.0 x 10-3L
L (Ca(OH) )
2)
2
HNO 3
units match!
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Limiting/Excess/ Reactant and Theoretical Yield Problems :
Potassium superoxide, KO2, is used in rebreathing gas masks to generate oxygen.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
a. How many moles of O2 can be produced from 0.15 mol KO2 and 0.10 mol H2O?
b. Determine the limiting reactant.
4KO2(s) + 2H2O(l)  4KOH(s) + 3O2(g)
0.15 mol
? moles
0.10 mol
Hide
Two starting
amounts?
Where do we
start?
one
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Try this problem (then check your answer):
Calculate the molarity of a solution prepared by dissolving 25.6 grams of
Al(NO3)3 in 455 mL of solution.
After you have
worked the
problem, click here
to see
setup answer
25.6 g mole
mol
 0.264
-3
213 g 455 x 10 L
L
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