The number of moles of $H_2$ required to produce $10$ moles of $NH_3$ in the following reaction is:
$a H_{2(g)} + b NO_{2(g)} \longrightarrow c NH_{3(g)} + d H_2O_{(g)}$

  • A
    $10$
  • B
    $20$
  • C
    $35$
  • D
    $53$

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