Answer:
M
g
+
2
H
C
l
⟶
M
g
C
l
2
+
H
2
(
g
)
Temperature,
T
=
0
∘
C
=
273
K
Pressure,
P
=
2
a
t
m
Volume of
H
C
l
,
V
1
=
275
m
L
Molarity of
H
C
l
,
M
1
=
0.725
M
Mass of Mg,
w
=
4.50
g
Moles of
H
C
l
,
m
1
=
M
1
×
V
1
m
1
=
M
1
×
V
1
=
0.725
×
0.275
=
0.2
m
o
l
e
Moles of Mg,
m
2
=
w
24.3
m
2
=
w
24.3
=
4.5
24.3
=
0.185
m
o
l
e
s
Explanation:
From the given reaction 1 mole of Mg reacts with 2 moles of HCl.
So, 0.185 moles of Mg requires 0.37 moles of HCl.
But we have only 0.2 moles of HCl available. Hence HCl is the limiting reagent.
From the given reaction, 2 moles of HCl produces 1 mole of hydrogen gas.
Therefore, 0.2 moles of HCl produces 0.1 moles of hydrogen gas.
We know,
P
V
=
n
R
T
V
=
n
R
T
P
=
0.1
m
o
l
×
0.0821
L
.
a
t
m
.
m
o
l
−
1
.
K
−
1
×
273
K
2
a
t
m
=
1.12
L
V
=
1.12
L
The volume of hydrogen gas produced is 1.12 liters.
