MCQ
${H_2}$ evolved at $STP$ on complete reaction of $27\, g$ of Aluminium with excess of aqueous $NaOH$ would be ............ $\mathrm{L}$
  • A
    $22.4$
  • B
    $44.8$
  • C
    $67.2$
  • $33.6$

Answer

Correct option: D.
$33.6$
d
(d) ${H_2}O + \mathop {Al}\limits_{27\,gm} + NaOH \to NaAl{O_2} + \mathop {\frac{3}{2}{H_2}\;\;\;\;}\limits_{\frac{3}{2} \times 22.4 = 33.6\,L} $

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$\begin{array}{*{20}{c}}
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  {C{H_3} - C{H_2} - CH - C{H_3}\xrightarrow[\Delta ]{{EtoNa}}\mathop {{\text{ }}Y}\limits_{\left( {{\text{major}}} \right)} } \\ 
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