Answer :
Answer:To determine the molecular formula of the gaseous hydrocarbon
�
A, we need to analyze the data provided and apply stoichiometric principles.
Initial Information:
Volume of hydrocarbon
�
A: 10 cm
3
3
Volume of oxygen: 50 cm
3
3
Volume of gas remaining after reaction: 40 cm
3
3
Volume of gas remaining after treatment with KOH: 30 cm
3
3
Reaction Details:
The hydrocarbon combusts completely in the presence of oxygen, forming carbon dioxide (
�
�
2
CO
2
) and water (
�
2
�
H
2
O):
CxHy
+
�
2
→
�
�
2
+
�
2
�
CxHy+O
2
→CO
2
+H
2
O
Volume Analysis:
The 40 cm
3
3
of gas remaining after cooling consists of
�
�
2
CO
2
and excess
�
2
O
2
.
After shaking with KOH,
�
�
2
CO
2
is absorbed, leaving 30 cm
3
3
of
�
2
O
2
.
Hence, the volume of
�
�
2
CO
2
produced is:
40
cm
3
−
30
cm
3
=
10
cm
3
40cm
3
−30cm
3
=10cm
3
Combustion Equation:
The balanced equation for the combustion of a hydrocarbon
CxHy
CxHy is:
CxHy
+
(
�
+
�
4
)
�
2
→
�
�
�
2
+
�
2
�
2
�
CxHy+(x+
4
y
)O
2
→xCO
2
+
2
y
H
2
O
Given that 10 cm
3
3
of hydrocarbon
�
A produces 10 cm
3
3
of
�
�
2
CO
2
:
�
=
1
x=1
Oxygen Calculation:
Hydrocarbon
�
A uses
(
�
+
�
4
)
(x+
4
y
) volumes of
�
2
O
2
.
From the initial 50 cm
3
3
of
�
2
O
2
, 30 cm
3
3
remains unreacted, meaning 20 cm
3
3
reacted with the hydrocarbon.
Thus:
�
+
�
4
=
2
x+
4
y
=2
Solving the Equations:
We have two equations:
�
=
1
x=1
1
+
�
4
=
2
1+
4
y
=2
Solving for
�
y:
�
4
=
1
4
y
=1
�
=
4
y=4
Molecular Formula:
The molecular formula of hydrocarbon
�
A is:
CH
4
CH
4
Thus, the molecular formula of hydrocarbon
�
A is CH
4
4
Explanation:
