Cheung, Anthony – Stoich 2 – Due: Oct 29 2006, midnight – Inst: McCord
1
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17
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time.
001
(part 1 of 1) 10 points
Balance the equation
? H
2
CO + ? H
2
O
→
? CH
4
+ ? O
3
.
The coefficients are
1.
3; 2; 1; 2.
2.
3; 3; 3; 2.
correct
3.
2; 1; 2; 1.
4.
3; 3; 2; 2.
5.
1; 2; 2; 3.
Explanation:
A balanced equation has the same num
ber of each kind of atom on both sides of the
equation.
We find the number of each kind
of atom using equation coefficients and com
position stoichiometry. For example, we find
there are 12 H atoms on the product side:
? H atoms = 3 CH
4
×
4 H
1 CH
4
= 12 H
The balanced equation is
3 H
2
CO + 3 H
2
O
→
3 CH
4
+ 2 O
3
,
and has 3 C, 12 H and 6 O atoms on each side.
The equation coefficients are 3, 3, 3, 2.
002
(part 1 of 1) 10 points
Consider the reaction
N
2
+ 3 H
2
→
2 NH
3
.
How much NH
3
can be produced from the
reaction of 74.2 g of N
2
and 14.0 moles of H
2
?
1.
5
.
62
×
10
24
molecules
2.
1
.
26
×
10
25
molecules
3.
1
.
59
×
10
24
molecules
4.
3
.
19
×
10
24
molecules
correct
5.
1
.
69
×
10
25
molecules
Explanation:
m
N
2
= 74.2 g
n
H
2
= 14.0 mol
First you must determine the limiting reac
tant:
? mol N
2
= 74
.
2 g N
2
×
1 mol N
2
28 g N
2
= 2
.
65 mol N
2
According to balanced equation, we need
3 mol H
2
1 mol N
2
.
We have
14
.
0 mol H
2
2
.
65 mol N
2
=
5
.
28 mol H
2
1 mol N
2
Therefore, H
2
is an excess and N
2
is limiting.
? molec NH
3
= 2
.
65 mol N
2
×
2 mol NH
3
1 mol N
2
×
6
.
022
×
10
23
NH
3
molec
1 mol NH
3
molec
= 3
.
19
×
10
24
molec NH
3
003
(part 1 of 1) 10 points
Consider the reaction
4 Fe(s) + 3 O
2
(g)
→
2 Fe
2
O
3
(s).
If 12.5 g of iron(III) oxide (rust) are pro
duced from 8.74 g of iron, how much oxygen
gas is needed for this reaction?
1.
None of these
correct
2.
8.74 g
3.
21.2 g
4.
7.5 g
5.
12.5 g
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Cheung, Anthony – Stoich 2 – Due: Oct 29 2006, midnight – Inst: McCord
2
Explanation:
m
iron
= 8.74 g
m
oxide
= 12.5 g
The balanced equation for the reaction tells
us that 4 mol Fe reacts with 3 mol O
2
to
produce 2 mol Fe
2
O
3
. We have two possible
starting points.
We know 12.5 g Fe
2
O
3
was
produced and that 8.74 g Fe was present at
the start of the reaction.
Choosing the 12.5 g of Fe
2
O
3
to start with,
first we convert to moles using the molar mass:
? mol Fe
2
O
3
= 12
.
5 g Fe
2
O
3
×
1 mol Fe
2
O
3
159
.
7 g Fe
2
O
3
= 0
.
0783 mol Fe
2
O
3
Now we use the mole ratio from the bal
anced equation to find moles O
2
needed to
produce 0.0783 mol Fe
2
O
3
.
? mol O
2
= 0
.
0783 mol Fe
2
O
3
×
3 mol O
2
2 mol Fe
2
O
3
= 0
.
117 mol O
2
We convert from moles to grams:
? g O
2
= 0
.
117 mol O
2
×
32 g O
2
1 mol O
2
= 3
.
744 g O
2
Starting with 8.74 g Fe and following the
same steps results in the same numerical an
swer.
004
(part 1 of 1) 10 points
How much O
2
is required for the complete
reaction of 36 g of C
3
H
6
to form CO
2
and
H
2
O?
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 Fakhreddine/Lyon
 Stoichiometry, mol, mol Co

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