Lin (cgl286) – Homework 8 – Lyon – (52370)
1
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This HW assignment is due Wednesday,
April 7, by 11PM.
001
10.0 points
An element X combines with oxygen to form
a compound of formula XO
2
. If 24.0 grams of
element X combines with exactly 16.0 grams
of O to form this compound, what is the
atomic weight of element X?
1.
24.0 amu
2.
48.0 amu
correct
3.
284 amu
4.
12.0 amu
5.
16.0 amu
Explanation:
m
X
= 24.0 g
m
O
= 16.0 g
We know from the formula XO
2
that one
mole of X combines with exactly 2 moles of
oxygen atoms.
We can convert 16 grams of
oxygen to moles using the atomic weight of
oxygen from the periodic table:
? mol O = 16 g O
×
1 mol O
16 g O
= 1 mol O
1 mol O would combine with exactly 1
/
2 mol
of X:
? mol X = 1 mol O
×
1 mol X
2 mol O
= 0
.
5 mol X
We were given that 16 g of O combines exactly
with 24 g X. We calculated that 16 g of O
(1 mol O) would combine with 0
.
5 mol X.
From this we can conclude that 0
.
5 mol X has
a mass of 24 g.
We can calculate the grams
per mole:
? molar mass X =
24 g X
0
.
5 mol X
=
48 g X
mol X
The molar mass of an element in g/mol is
numerically equal to the atomic mass in amu.
Therefore, the atomic mass of X is 48 amu.
002
10.0 points
27 g of Al reacts with 48 g of S to produce
Al
2
S
3
, the only product, with no reactant left
over.
How much Al
2
S
3
would be produced
from 78 g of Al and 101 g of S?
1.
158 g
correct
2.
179 g
3.
281 g
4.
122 g
5.
217 g
6.
27 g
Explanation:
m
ini
,
Al
= 27 g
m
ini
,
S
= 48 g
m
Al
= 78 g
m
S
= 101 g
By the law of conservation of mass, 27 g of
Al and 48 g of S would produce 75 g of Al
2
S
3
,
if no Al and S are remaining at the end of the
reaction.
By the law of definite proportions
then, the mass ratio of product Al
2
S
3
to either
reactant (Al or S) is unchanging.
However,
the mass ratio of Al to S is different in the
reaction at hand than in the ideal reaction, so
one of the reactants must be limiting.
If we assume that Al is limiting, the following
relationship is true:
75 g Al
2
S
3
27 g Al
=
x
g Al
2
S
3
78 g Al
x
= 217 g Al
2
S
3
Then if we assume that S is limiting,
75 g Al
2
S
3
48 g S
=
y
g Al
2
S
3
101 g S
y
= 158 g Al
2
S
3
.
Since the number of grams of Al
2
S
3
is
smaller when we assume S is the limiting
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Lin (cgl286) – Homework 8 – Lyon – (52370)
2
reagent, S really is the limiting reagent and
only 158 g of Al
2
S
3
is formed.
003
10.0 points
Consider the reaction
CaCN
2
+ 3 H
2
O
→
CaCO
3
+ 2 NH
3
,
which has a 45.6% yield.
How much H
2
O is
needed to produce 16.3 moles of CaCO
3
?
1.
107 g
2.
401 g
3.
48.9 g
4.
22.3 g
5.
1930 g
correct
6.
880 g
Explanation:
Due to the % yield, the 16.3 moles of CaCO
3
is the ACTUAL yield. We must find the THE
ORETICAL yield (which is always greater):
(16
.
3 mol CaCO
3
)
×
100%
45
.
6%
= 35
.
7456 mol CaCO
3
Find the moles of water needed:
(35
.
7456 mol CaCO
3
)
×
3 mol H
2
O
1 mol CaCO
3
= 107
.
237 mol H
2
O
The formula weight of water is
2(1
.
0079 g
/
mol) + 16
.
0 g
/
mol
= 18
.
0158 g
/
mol
so the mass of water needed is
(107
.
237 mol H
2
O)
×
18
.
0158 g H
2
O
1 mol H
2
O
= 1931
.
96 g H
2
O
004
10.0 points
Consider the reaction
N
2
+ 3 H
2
→
2 NH
3
which has a 45.6% yield.
28.0 g of N
2
and
9.04 g of H
2
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 Spring '09
 Lyon
 mol, ml, Sodium hydroxide

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