Balancing Chemical Equations
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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