propane
propene
propyne
propanone
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CxHy+(x+y/4)O2 --------- xCO2+y/2H2O
from the equation x=3
3+y/4=4
multiply both sides by 4
3×4+y×4/4=4×4
12+y=16
y=16-12=4
y=4
C3H4+(3+4/4)O2 ---------- 3CO2+4/2H2O
C3H4+4O2 --------- 3CO2+2H2O
C3H4.........is propyne
The selected ans is correct
Option C
in d product we have 3C
and. 2H2 >4H
therefore we can say we have
C3.
and
H4
which is............. C3H4 ....,...> Propyne
Count the Carbons (C): On the right, you have 3CO2. That is 3 Carbons. So, x = 3.
Count the Hydrogens (H): On the right, you have 2H_2O. That is 2x2 = 4Hydrogens. So, y = 4.
Identify the molecule: The formula is C3H4.shey una get am now.
just use law of conservation of mass proposed by Lavoisier
Step 1: Use CO₂ and H₂O to find number of C and H atoms
From the right side of the equation:
3CO₂ → means the hydrocarbon has 3 carbon atoms (C₃)
2H₂O → means it has 4 hydrogen atoms (since each H₂O has 2 H, so 2×2 = 4 H)
So, the hydrocarbon is C₃H₄
Step 2: Which option is C₃H₄?
Let's go through the options:
A. Propane = C₃H₈ 
B. Propene = C₃H₆
C. Propyne = C₃H₄ 
D. Propanone = C₃H₆O
Correct Answer: C. Propyne
Memory Tip:
"-yne" in propyne = alkyne, meaning it has a triple bond and fewer hydrogens than alkanes (like propane) or alkenes (like propene).
CₙH₂ₙ₋₂ = General formula for alkynes (like C₃H₄).
