Write the reaction of aluminum with caustic soda and determine how many moles of $H_2$ gas will be produced at $STP$ from $54 \ g$ of $Al$. (in $mol$)

  • A
    $1$
  • B
    $2$
  • C
    $3$
  • D
    $4$

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Volume of $3 \ M \ NaOH$ (formula weight $40 \ g \ mol^{-1}$) which can be prepared from $84 \ g$ of $NaOH$ is $ . . . . . . \times 10^{-1} \ dm^3$.

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