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Pertanyaan

Akrilonitril diproduksi dengan mereaksikan propilen, amonia, dan oksigen seperti pada reaksi
C3H6+NH3+1,5O2----->C3H3N+3H2O
umpan reaktor berisi 10%propilen, 12% amonia, 78% udara. Tentukan jika konversi limiting reactant hanya 30% berapa rasio{mol akrilonitril / mol NH3 umpan}t

1 Jawaban

  • C3H6 + NH3 + 3/2 O2 -> C3H3N + 3H2O
    basis perhitungan umpan masuk 100mol
    Masuk :
    C3H6 = 10 mol
    NH3 = 12 mol
    O2 = 21/100 x 78 = 16,38 mol
    N2 = 61,62 mol
    Bereaksi :
    C3H6 = 10x30% = 3 mol
    NH3 = 3 mol
    O2 = 1,5 x 3 mol = 4,5 mol
    C3H3N = 3 mol
    H2O = 9mol
    Sisa
    C3H6 = 7 mol
    NH3 = 9 mol
    O2 = 11,88 mol
    N2 = 61,62 mol
    C3H3N = 3 mol
    H2O = 9 mol

    rasio = 3mol/12mol = 0,25

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