Transcript Balancing Chemical Equations

Titrations
Viatmin C is both an acid and a reducing
agent. One method of determining the
amount of Vit C in a sample is to carry out the
following reaction via a titration:
2H+(aq) + C6H8O8(aq) + 2Br2(aq)
4HBr(aq) + C6H6O6 (aq) + 2H2O
The vitamin C in a 1.00 g chewable tablet
requires 28.28 mL of 0.102 M Br2 solution for
titration to the equivalence point. How many
grams of vitamin C (208 g/mol) are contained
in the tablet?
mols
C6H8O8
g
C6H8O8
0.02828 L
x  x
1
1 mols C6H8O8

2 mol Br2
mols Br2
0.102 mols Br2

L
x
208 g C6H8O8

mol C6H8O8
=
0.300 g C6H8O8
Viatmin C is both an acid and a reducing
agent. One method of determining the
amount of Vit C in a sample is to carry out the
following reaction via a titration:
2H+(aq) + C6H8O8(aq) + 2Br2(aq)
4HBr(aq) + C6H6O6 (aq) + 2H2O
The vitamin C in a 1.00 g chewable tablet
requires 28.28 mL of 0.102 M Br2 solution for
titration to the equivalence point. How many
grams of vitamin C (208 g/mol) are contained
in the tablet?
mols
C6H8O8
g
C6H8O8
0.02828 L
x  x
1
1 mols C6H8O8

2 mol Br2
mols Br2
0.102 mols Br2

L
x
208 g C6H8O8

mol C6H8O8
=
0.300 g C6H8O8
A 50.00 mL sample of solution containing
Fe2+ is titrated with 0.0216 M KMnO4
solution. The solution required 20.65 mL of
KMnO4 solution to oxidize all of the Fe2+
ions to Fe3+ by the reaction:
8H+(aq) + MnO41-(aq) + 5Fe2+(aq) 
Mn2+(aq) +5Fe3+(aq) + 4H2O
1
mols
KMnO4
2
0.0216 mols
KMnO4

L
x
mols
MnO41-
mols
Fe2+
0.02065 L
x  x
1
5 mols Fe2+
 x
1 mol MnO41-
1

0.0500 L
M
Fe2+
1 mols MnO41
1 mol KMnO4
=
0.0446 M Fe2+
A 50.00 mL sample of solution containing
Fe2+ is titrated with 0.0216 M KMnO4
solution. The solution required 20.65 mL of
KMnO4 solution to oxidize all of the Fe2+
ions to Fe3+ by the reaction:
8H+(aq) + MnO41-(aq) + 5Fe2+(aq) 
Mn2+(aq) +5Fe3+(aq) + 4H2O
1
mols
KMnO4
2
0.0216 mols
KMnO4

L
x
mols
MnO41-
mols
Fe2+
0.02065 L
x  x
1
5 mols Fe2+
 x
1 mol MnO41-
1

0.0500 L
M
Fe2+
1 mols MnO41
1 mol KMnO4
=
0.0446 M Fe2+
The Limiting Reagent
relative to the other reactant(s) of a chemical
reaction, this reactant is present in less than
the stoichiometrically equivalent amount
i.e. you have less than you need to fully react
with other reactants.
determines,i.e. limits, the quantity of
product(s) that will be obtained.
is totally consumed during the chemical reaction.
Reactants other than the limiting reagent are
in excess (i.e. in excess of that amount
required for stoi- chiometric equivalence with
the limiting reagent). Some quantity of this
(these) reactant(s) will remain after the
reaction is complete.
Example
• Desire
• Provided with:
and
Which,
or
, is the
“limiting” supply?
CaC2(s) + 2H2O
Ca(OH)2(aq) +C2H2(g)
How many grams of C2H2(g) will be formed from
the reaction of 24.0 g CaC2(s) and 18.0 g H2O ?
Empirical Formula
The simplest wholenumbered
(3)
ratio
(2)
of numbers of mols of
atoms
(1)
in one mol of a
compound.
Mo(CO)x(s)
Mo(s) + xCO(g)
When a 2.200 g of Mo(CO)x is heated it
decomposes producing 0.7809 g of Mo(s)
and gaseous CO. The CO gas was found
to occupy a volume of 1.2345 L at a
temperature of 31.50 oC and a pressure of
751.1 mm Hg. What is the value of x?
R= 0.0821 L-atm/mol K
R = 62,400 mL-mm/mol K
A compound with the general formula CxHy was
vaporized and, at 0.00 oC and 760 mm Hg, was
found to have a density of 5.0996 g per L. In a
separate determination the elemental composition
of the compound was found to be 84.118 % C and
15.882 % H.
(1). Calculate the molar mass of the compound.
(2). Calculate the empirical formula of the
compound
(3). What is the molecular formula of the
compound?
R= 0.0821 L-atm/mol K
R = 62,400 mL-mm/mol K
C(s) + 2 H2O(l)
CO2 + 2 H2(g)
How many grams of hydrogen gas are produced
when 18 g C reacts with 27 g H2O?
C(s) + 2H2O  CO2(aq)
+2H2(g)
From the balanced chemical equation:
1 mol C
0.50 mols C
or

2 mols H2O

mol H2O
Provided:
mol C
18.0g C x
= 1.5 mol C

12.0 g C
27g H2O x
mol H2O

18.0 g H2O
= 1.5 mol H2O
Compare:
Provided
1 mol C

1 mols H2O
vs
From
Reaction stoichiometry
0.50 mols C

1 mol H2O
Conclusion:
H2O is in the limiting reagent, C is in excess
mol H2O(l)
2 mol H2
27.0 g H2 O (l) x  x 
18.0 g H2O(l)
2 mols H2O(l)
2.02 g H2
x 
mol H2(g)
=
3.03 g H2
How many grams of precipitate will be
formed when 308.6 mL of 0.324 M Al2(SO4)3
is poured into 432. mL of 1.157 M NaOH?
Al2(SO4)3(aq)
3 Na2SO4 (aq)
+
6
NaOH
+ 2 Al(OH)3 (s)
How many grams of N2O(g) will be produced
when 14.0 g N2(g) reacts with 30.0 g H2O(g)?
N2(g) + H2O(g)
N2O(g) + NH3(l)
How many grams H2(g) will be formed when
2.16 g Al react with 2.92 g HCl (in aqueous
solution)?
2 Al(s) + 6HCl(aq)

2 AlCl3(aq) + 3H2(g)
Al4C3(s) + H2O
Al(OH)3(s) +
CH4(g)
How many grams of CH4(g) will be formed from
the reaction of 14.4 g Al4C3(s) and 18.0 g H2O ?
Al4C3(s) + H2O 
Al(OH)3(s) +
CH4( g)
Which is the limiting reagent?
From the balanced chemical reaction:
1 mol Al4C3(s)
0.0833 mol Al4C3(s)

12 mols H2O
=

1 mols H2O
Provided:
mol Al4C3
14.4 g All4C3(s) x  = 0.100 mol Al4C3
144 g Al4C3
mol H2O
18g H2O x  = 1.0 mol H2O
18.0 g H2O
Compare:
From
Reaction stoichiometry
Provided
0.100 mol Al4C3

1.0 mol H2O
0.0833 mol Al4C3
vs

1.0 mol H2O
Al4C3(s) is in excess, , H2O is the limiting reagent
Quantity of the product that will be obtained will
be determined by the quantity of the limiting
reagent provided.
1 mol H2O (l)
3 mol CH4
16.0 g CH4
18.0 g H2O (l) x  x  x 
18.0 g H2O (l) 12 mols H2O (l) mol CH4
= 4.0 g CH4
Which is the limiting reagent?
From the balanced chemical reaction:
1 mol Al4C3(s)
12 mols H2O
Now, how many molsAl4C3(s) are needed to react
with the number of mols of H2O provided?
1 mol H2O(s)
18.0 g H2O(s) x
x
18 g H2O(s)
1 molsAl4C3(s)
=
12 mols H2O (l)
= 0.083 molsAl4C3(s) (needed)
18.0g H2O requires 0.083 mols Al4C3(s) for
complete reaction.
14. 4 g Al4C3(s) is provided:
1 mol Al4C3(s)O
14.4 g Al4C3(s) x
= 0.10 mols Al4C3(s)
144 g Al4C3(s)
0.10 mols Al4C3(s) (provided) > 0.083 mols Al4C3(s) H2O
(needed )
Conclusion:
Al4C3(s) is in excess, H2O is the limiting reagent
Quantity of the product that will be obtained will
be determined by the quantity of the limiting
reagent provided.
1 mol H2O (l)
3 mol CH4
16.0 g CH4
18.0 g H2O (l) x —————— x ————— x —————
18.0 g H2O (l) 12 mols H2O (l) mol CH4
= 4.0 g CH4