Nme.docx

1. Basis = 1 jam operasi Udara masuk = 6000 kg BM Udara = 0.79 (28) + 0.21 (32) = 28.84 6000

Udara = 28.84 = 208.044 𝑘𝑔𝑚𝑜𝑙 O2 dari udara = 0.21 x 208.044 kgmol = 43.689 kgmol N2 dari udara = 0.79 x 208.044 kgmol = 164.354 kgmol 100

O2 teoritis = 100+90 𝑥 43.689 𝑘𝑔𝑚𝑜𝑙 = 22.994 𝑘𝑔𝑚𝑜𝑙

Misal campuran bijih besi = 100x kg FeS2 = 90x kg

 FeS2 =

ZnS = 5x kg

ZnS =

90𝑥 120

5𝑥 97

𝑥 1 𝑚𝑜𝑙 = 0.75𝑥 𝑘𝑔𝑚𝑜𝑙

𝑥 1 𝑚𝑜𝑙 = 0.05𝑥 𝑘𝑔𝑚𝑜𝑙

Reaksi I dan III  

4 FeS2 + 11 O2  2 Fe2O3 + 8SO2 0.75x

11 4

ZnS 0.05x

+ 1.5 O2  ZnO + SO2 1.5x0.05 = 0.077x

𝑥 0.75𝑥 = 2.062

O2 untuk reaksi

= O2 teoritis

2.062 x + 0.077x

= 22.994 kgmol

X

= 10.747 kgmol

BM Campuran FeS2 dan ZnS = 0.90 x 120 + 0.05 x 97 = 108 + 4.85 = 112.85 Berat bijih besi non inert (95%) = 10.747 x 112.85 kg = 1212.832 kg Dalam bijih besi total =

100 95

𝑥 1212.832𝑘𝑔 = 1276.666 𝑘𝑔

Inert dalah bijih besi = 1276.666 kg – 1212.832 kg = 63.833 kg FeS2 = 0.90 x 1212.823 kg = 1091.541 kg x ZnS = 0.05 x 1212.823 kg = 60.641 x

1 𝑚𝑜𝑙 97

1 𝑚𝑜𝑙 120

= 9.096 kgmol

= 0.625 𝑘𝑔𝑚𝑜𝑙

FeS2 yang bereaksi = 0.9 x 9.096 kgmol = 8.186 kgmol FeS2 sisa = 0.1 x 9.096 kgmol = 0.909 kgmol 4FeS2 + 11 O2  2Fe2O3 + 8SO2 FeS2 = 0.8 x 8.186 kgmol = 6.549 kgmol O2 = 18.009 kgmol Fe2O3 = 3.274 kgmol SO2 = 13.098 kgmol 4FeS2 + 15 O2  2Fe2O3 + 8SO3 FeS2 = 0.2 x 8.186 kgmol = 1.637 kgmol O2 = 6.14kgmol Fe2O3 = 0.818 kgmol SO3 = 3.2746 kgmol  Reaksi I ZnS yang bereaksi = 0.9 x 0.6225 kgmol = 0,560 kgmol ZnS sisa = 0.1 .6225 kgmol = 0.062 kgmol ZnS + 2O2  ZnO + SO2 ZnS = 0.8 0.560 kgmol = 0.448 kgmol O2 = 0.672 kgmol ZnO = SO2 =0.448 kgmol 

Reaksi II

ZnS + 2O2  ZnO + SO3 ZnS = 0.2 x 0.50 kgmol = 0.112 kgmol O2 = 0.224 kgmol ZnO = SO3 = 0.112 kgmol Total kebutuhan O2 reaksi = (18.01 +6.14 + 0.672 + 0.224)kgmol = 25.046 kgmol

O2 dari udara = 43.689 kgmol Produk FeS2 sisa = 0.909 kgmol ZnS = 0.062 kgmol O2 sisa = 43.689-25.046 = 18.822 kgmol SO2 total = 13.098 + 0.448 = 13.546 kgmol SO3 total = 3.274 + 0.112 = 3.386 kgmol Fe2O3 total = 3.274 + 0.818 = 4.093 kgmol ZnO total = 0.448 + 0.112 = 0.560 kgmol N2 = 164.354 kgmol Inert = 68.833 kgmol INPUT Bijih besi FeS2 ZnS Inert O2 N2 Total

Kgmol 9.096 0.622

BM 120 97

43.689 164.354

32 28

Kgmol 13.546 3.386 18.822 164.354 4.093 0.560 0.9096 0.062

BM 64 80 32 28 160 81 120 97

kg 1091.520 60.382 63.833 1398.048 460.9120 7215.695

OUTPUT Bijih besi SO2 SO3 O2 sisa N2 Fe2O3 ZnO FeS2 sisa ZnS sisa Inert Total

kg 866.963 270.9280 602.332 4601.912 654.896 45.368 109.152 6.033 63.833 7221.417

CONVERTER Reaksi : SO2 + 0.5 O2  SO3 SO2 total = 13.546 kgmol SO2 yang bereaksi = 0.95 x 13.546 kgmol = 12.657 kgmol Kebutuhan O2 = 0.5 x 12.657 kgmol = 6.328 kgmol SO3 hasil reaksi = 12.657 kgmol Produk SO3 total = 12.657 + 3.386 kgmol = 16.043 kgmol SO2 sisa = 0.673 kgmool O2 sisa = 18.822 – 6.328 12.494 kgmol N2 = 164.354 kgmol Neraca massa pada converter INPUT Bijih besi SO2 SO3 O2 sisa N2 Total

Kgmol 13.546 3.386 18.822 164.354

BM 64 80 32 28

kg 866.963 270.928 602.332 4601.912 6342.136

Kgmol 0.673 16.043 12.494 164.354

BM 64 80 32 28

kg 43.347 1283.480 399.820 4601.912 6328.568

OUTPUT Bijih besi SO2 sisa SO3 O2 sisa N2 Total

Neraca Overall INPUT Bijih Besi

Kgmol

BM

kg

FeS2 ZnS Inert O2 N2 Total

8.096 0.622

120 97

43.689 164.354

32 28

1091.520 60.382 63.833 1398.048 4601.912

Kgmol 13.546 3.386 18.822 164.354 4.093 0.560 0.996 0.065

BM 64 80 32 28 160 81 120 91

OUTPUT Bijih besi SO2 SO3 O2 sisa N2 Fe2O3 ZnO FeS2 sisa ZnS sisa Total

2. Basis = 100.000 kg NaOH murni = 0.8 x 100.000 = 80.000 kg H2O dalam NaOH = 0.2 x 100.000 = 20.000 kg BM NaOH = 40 gr/mol Mol NaOH =

80.000.000 40

= 2.000.000 𝑚𝑜𝑙 = 20.000 𝑘𝑔𝑚𝑜𝑙

NaOH berlebih = 80% = 20.000 kgmol NaOH =

100% 100%+80%

x 20.000 = 10.000 kgmol

Reaksi : 2 NaOH + H2SO4 10.000 0.5x10.000 =5.000 H2SO4 teoritis = 5000



H2SO4 96% yang digunakan =

100% 96%

Na2SO4 + 2H2O 5000 10.000

𝑥 H2SO4 teoritis

kg 866.963 270.928 602.332 4601.912 654.896 45.368 109.152 6.033 6342.136

=

100% 96%

𝑥 5000

= 5208 𝑘𝑔𝑚𝑜𝑙 H2O dalam H2SO4 =

4 𝑥5208 100

= 201.12 𝑘𝑔𝑚𝑜𝑙 80

Derajat Kesempurnaan reaksi 80%  100 𝑥5000 = 4000𝑘𝑔𝑚𝑜𝑙 

Reaksi I :

90 100

𝑥 4000 = 3600 𝑘𝑔𝑚𝑜𝑙

2NaOH + H2SO4  7200 3600 

Na2SO4 + 2H2O 3600 7200

10

Reaksi II : 100 𝑥4000 = 400 𝑘𝑔𝑚𝑜𝑙 NaOH + H2SO4  400 400

Sisa 20%  KOMPOSISI H2SO4 sisa Na2SO4 NaHSO4 H2O Total

𝟐𝟎 𝒙 𝟓𝟎𝟎𝟎 𝟏𝟎𝟎

NaHSO4 + H2O 400 400 = 𝟏𝟎𝟎𝟎𝒌𝒈𝒎𝒐𝒍 PRODUK 1000kgmol 3600 kgmol 400kgmol 7600 kgmol 12.600kgmol