A) \[{{C}_{4}}{{H}_{10}}\]
B) \[{{C}_{3}}{{H}_{6}}\]
C) \[{{C}_{3}}{{H}_{8}}\]
D) \[{{C}_{4}}{{H}_{8}}\]
Correct Answer: C
Solution :
Volume of \[{{N}_{2}}\]in air = 375 × 0.8 = 300 ml volume of \[{{O}_{2}}\]in air = 375 × 0.2 = 75 ml \[{{C}_{x}}{{H}_{y}}+\left( x+\frac{y}{4} \right){{O}_{2}}\xrightarrow[{}]{{}}xC{{O}_{2}}(g)+\frac{y}{2}{{H}_{2}}O(\ell )\] \[15\,ml\,15\left( x+\frac{y}{4} \right)\] 0 0 15 - After combustion total volume \[330={{V}_{{{N}_{2}}}}+{{V}_{C{{O}_{2}}}}\] \[330=300+15x\] \[x=2\] Volume of \[{{O}_{2}}\]used \[15\left( x+\frac{y}{4} \right)=75\] \[x+\frac{y}{4}=5\] \[y=12\] So hydrocarbon is\[={{C}_{2}}{{H}_{12}}\] none of the option matches it therefore it is a BONUS. Alternatively \[{{C}_{x}}{{H}_{y}}+\left( x+\frac{y}{4} \right){{O}_{2}}\xrightarrow[{}]{{}}xC{{O}_{2}}+\frac{y}{2}{{H}_{2}}O(\ell )\] 15 \[15\left( x+\frac{y}{4} \right)\] 0 0 15x - Volume of \[{{O}_{2}}\]used \[15\left( x+\frac{y}{4} \right)=75\] \[x+\frac{y}{4}=5\] If further information (i.e., 330 ml) is neglected, option (3) only satisfy the above equation.
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