MCQ
$4.3\ gm$ of an alkane is burnt in sufficient oxygen. The $CO_2$ formed reacts completely with $300\ ml$, $2\,N\ NaOH$ solution producing $Na_2CO_3$. The alkane should be
- A$C_3H_8$
- B$C_{12}H_{26}$
- ✓$C_6H_{14}$
- D$C_2H_6$
✓
Answer
Correct option: C.
$C_6H_{14}$
c
$\mathop {{{\text{C}}_{\text{n}}}{{\text{H}}_{2{\text{n}} + 2}}}\limits_{\frac{{0.3}}{{\text{n}}}{\mkern 1mu} \,{\text{mole}}} + \frac{{(3{\text{n}} + 1)}}{2}{{\text{O}}_2} \to \mathop {{\text{nC}}{{\text{O}}_2}}\limits_{0.3\,{\mkern 1mu} {\text{mole}}} + ({\text{n}} + 1){{\text{H}}_2}{\text{O}}$
$\mathop {{{\text{C}}_{\text{n}}}{{\text{H}}_{2{\text{n}} + 2}}}\limits_{\frac{{0.3}}{{\text{n}}}{\mkern 1mu} \,{\text{mole}}} + \frac{{(3{\text{n}} + 1)}}{2}{{\text{O}}_2} \to \mathop {{\text{nC}}{{\text{O}}_2}}\limits_{0.3\,{\mkern 1mu} {\text{mole}}} + ({\text{n}} + 1){{\text{H}}_2}{\text{O}}$
$\mathop {{\text{C}}{{\text{O}}_2}}\limits_{0.3\,mole} + \mathop {2{\text{NaOH}}}\limits_{0.6\,mole} \to {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + {{\text{H}}_2}{\text{O}}$
$\frac{0.3}{n}=\frac{4.3}{(14 n+2)} \Rightarrow n=6$
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