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Hydrocarbons question

2016 · 9 Apr · Shift 1 · Q6
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Hydrocarbons question

2016 · 9 Apr · Shift 1 · Q6

JEE MainChemistryHydrocarbonsMCQ+4 / −1
5 L of an alkane requires 25 L of oxygen for its complete combustion. If all volumes are measured at constant temperature and pressure, the alkane is :
  1. A
    Ethane
  2. B
    Propane
  3. C
    Butane
  4. D
    Isobutane
View written solutionFree

Correct answer: B

  1. Write the general formula of an alkane

For an alkane, Alkane=CnH2n+2\text{Alkane} = C_nH_{2n+2}Alkane=Cn​H2n+2​

  1. Write the combustion reaction

Complete combustion of CnH2n+2C_nH_{2n+2}Cn​H2n+2​ is: CnH2n+2+3n+12O2→nCO2+(n+1)H2OC_nH_{2n+2} + \frac{3n+1}{2}O_2 \rightarrow nCO_2 + (n+1)H_2OCn​H2n+2​+23n+1​O2​→nCO2​+(n+1)H2​O

So, 1 volume of alkane requires 3n+12\frac{3n+1}{2}23n+1​ volumes of O2O_2O2​.

  1. Use the given volume ratio

Given:

  • Alkane volume =5 L= 5\,L=5L
  • Oxygen volume =25 L= 25\,L=25L

Thus, oxygen required per 1 volume of alkane is: 255=5\frac{25}{5} = 5525​=5

So, 3n+12=5\frac{3n+1}{2} = 523n+1​=5

  1. Solve for nnn

3n+1=103n+1 = 103n+1=10 3n=93n = 93n=9 n=3n = 3n=3

Therefore, the alkane is: C3H8C_3H_8C3​H8​ which is propane.

  1. Check options
  • A: Ethane =C2H6= C_2H_6=C2​H6​ ❌
  • B: Propane =C3H8= C_3H_8=C3​H8​ ✅
  • C: Butane =C4H10= C_4H_{10}=C4​H10​ ❌
  • D: Isobutane =C4H10= C_4H_{10}=C4​H10​ ❌

Hence, the correct option is B.

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