The mass of $N_2F_4$ produced by the reaction of $2.0 \ mol$ of $NH_3$ and $8.0 \ mol$ of $F_2$ is $0.5 \ mol.$ What is the per cent yield?
$2NH_3 + 5F_2 \to N_2F_4 + 6HF$

  • A
    $79.0$
  • B
    $71.2$
  • C
    $84.6$
  • D
    $50.0$

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