Answer:0.0324
g Al
Explanation:We're asked to find the number of grams of
Al
that reacted, given some
H
2
(
g
)
product characteristics.
Let's first write the chemical equation for this reaction:
2
Al
(
s
)
+
6
HCl
(
a
q
)
→
2
AlCl
3
(
a
q
)
+
3
H
2
(
g
)
The total pressure of the gaseous system is given as
751
mm Hg
, and the partial pressure of water vapor is
26.8
mm Hg
at
27
o
C
. The pressure of hydrogen gas is thus
P
total
=
P
H
2
O
+
P
H
2
P
H
2
=
751
mm Hg
−
26.8
mm Hg
=
724
mm Hg
This pressure in atmospheres is
724
mm Hg
(
1
l
atm
760
mm Hg
)
=
0.953
atm
We'll now use the ideal gas equation to find the number of moles of
H
2
formed:
(
T
=
27
o
C
+
273
=
300
K
)
n
=
P
V
R
T
=
(
0.953
atm
)
(
0.0465
L
)
(
0.082057
L
∙
atm
mol
∙
K
)
(
300
K
)
=
0.00180
mol H
2
(volume converted to liters here)
Using the coefficients of the chemical equation, we'll now find the relative number of moles of
Al
that react:
0.00180
mol H
2
(
2
l
mol Al
3
mol H
2
)
=
0.00120
mol Al
Lastly, we'll use the molar mass of aluminum (
26.98
g/mol
) to find the number of grams that reacted:
0.00120
mol Al
(
26.98
l
g Al
1
mol Al
)
=
0.0324
g Al
Thus,
0.0324
grams of aluminum
reacted.