Consider the reaction
$4 HNO_{3(\ell)} + 3 KCl_{(s)} \rightarrow Cl_{2(g)} + NOCl_{(g)} + 2 H_{2}O_{(g)} + 3 KNO_{3(s)}$
The amount of $HNO_{3}$ required to produce $110.0 \ g$ of $KNO_{3}$ is $...... \ g$.
(Given: Atomic masses of $H, O, N$ and $K$ are $1, 16, 14$ and $39$ respectively.)

  • A
    $32.2$
  • B
    $69.4$
  • C
    $91.5$
  • D
    $162.5$

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