Lembar Perhitungan Reagen
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LEMBAR PERHITUNGAN REAGEN ETIL ASETAT 0,1 N Etil Asetat M = massa BM × 1000 V × ekuivalen × % kadar 0,1 = masssa 88 gr mol × 1000 2000 × 1 × 99 % Massa = 17,77 gram ,ρ=0 , 887 gr ml V asetat = masssa ρ =20,03 ml 0,2 N Etil Asetat M = massa BM × 1000 V × ekuivalen × % kadar 0,2 = masssa 88 gr mol × 1000 2000 × 1 × 99 % Massa = 35,55 gram ,ρ=0 , 887 gr ml V asetat = masssa ρ =40,07 ml 0,3 N Etil Asetat M = massa BM × 1000 V × ekuivalen × % kadar 0,3 = masssa 88 gr mol × 1000 2000 × 1 × 99 % Massa = 53,33 gram ,ρ=0 , 887 gr ml V asetat = masssa ρ =60,12 ml
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Transcript of Lembar Perhitungan Reagen
LEMBAR PERHITUNGAN REAGENETIL ASETAT 0,1 N Etil AsetatM = 0,1 = Massa = 17,77 gramV asetat =
0,2 N Etil AsetatM = 0,2 = Massa = 35,55 gramV asetat =
0,3 N Etil AsetatM = 0,3 = Massa = 53,33 gramV asetat =
0,1 N NaOHM = 0,1 = Massa = 8,16 gram
0,1 N HClM = 0,1 = Massa = 7,3 gramV HCl =
LEMBAR PERHITUNGAN
VARIABEL IPROSES BATCHPROSES KONTINYU
tVtitran (ml)Ca, V1N1 = V2N2tVtitranCa, V1N1 = V2N2
02,35 . x = 2,3 . 0,1x = 0,04601,85 . x = 1,8 . 0,1x = 0,036
12,25. x = 2,2 . 0,1x = 0,04411,65 . x = 1,6 . 0,1x = 0,032
22,15 . x = 2,1 . 0,1x = 0,04221,55 . x = 1,5 . 0,1x = 0,030
31,95 . x = 1,9 . 0,1x = 0,03831,45 . x = 1,4 . 0,1x = 0,028
425 . x = 2 . 0,1x = 0,04041,55 . x = 1,5 . 0,1x = 0,030
VARIABEL IIPROSES BATCHPROSES KONTINYU
tVtitran (ml)Ca, V1N1 = V2N2tVtitran (ml)Ca, V1N1 = V2N2
01,75 . x = 1,7 . 0,1x = 0,03401,55 . x = 1,5 . 0,1x = 0,030
11,65. x = 1,6 . 0,1x = 0,03211,55 . x = 1,6 . 0,1x = 0,030
21,55 . x = 1,5 . 0,1x = 0,03021,45 . x = 1,5 . 0,1x = 0,028
31,65 . x = 1,4 . 0,1x = 0,02831,45 . x = 1,4 . 0,1x = 0,028
VARIABEL IIIPROSES BATCHPROSES KONTINYU
tVtitran (ml)Ca, V1N1 = V2N2tVtitranCa, V1N1 = V2N2
01,75 . x = 1,7 . 0,1x = 0,03401,45 . x = 1,4 . 0,1x = 0,028
11,65. x = 1,6 . 0,1x = 0,03211,25 . x = 1,2 . 0,1x = 0,024
21,65 . x = 1,6 . 0,1x = 0,03221,35 . x = 1,3 . 0,1x = 0,026
31,55 . x = 1,3 . 0,1x = 0,03031,35 . x = 1,3 . 0,1x = 0,026
PERHITUNGAN ORDE REAKSI
VARIABEL ICa = CNaOH Ca0 = CNaOH0 = 0,1 NCb = CEtil Asetat = 0,1 NData Perhitungan Grafik Trial Orde 1tCa
00,0460,776
10,0440,821
20,0420,867
30,0380,967
40,0400,916
Grafik Data Trial Orde 1
Data perhitungan Grafik Trial Orde II Ca = CbtCa
00,04621,73910
10,04422,72710
20,04223,80910
30,03826,31510
40,0402510
Grafik Data Trial Orde 2
VARIABEL IICa = CNaOH Ca0 = CNaOH0 = 0,1 NCb = CEtil Asetat = 0,1 NData Perhitungan Grafik Trial Orde 1tCa
00,0341,078
10,0321,139
20,0301,203
30,0321,139
Grafik Data Trial Orde 1
Data Perhitungan Grafik Trial Orde II Ca CbCb = Cb0 Ca0XaXa = 1 - tCaXaCb
00,0340,660,1341,3710,693
10,0320,680,1321,4170,693
20,0300,700,1301,4660,693
30,0280,720,1281,5190,693
Grafik Data Trial Orde II
VARIABEL IIICa = CNaOH Ca0 = CNaOH0 = 0,1 NCb = CEtil Asetat = 0,3 NData Perhitungan Grafik Trial Orde ItCa
00,0341,078
10,0321,139
20,0321,139
30,0301,203
Grafik Data Trial Orde 1
Data Perhitungan Grafik Trial Orde II Ca CbCb = Cb0 Ca0XaXa = 1 - tCaXaCb
00,0340,660,2341,9281,098
10,0320,680,2321,9811,098
20,0320,680,2321,9811,098
30,0300,700,2302,0361,098
Grafik Data Trial Orde II
PERHITUNGAN NILAI k LAJU REAKSI ORDE II VARIABEL I. y = mx + cy = 1,010x + 21,89
m1 = m2 =
m3 =
m4 =
m5 =
Sehingga, = 1,012
VARIABEL II. y = mx + c y = 0,049x + 1,369
m1 =
m2 =
m3 =
m4 =
Sehingga, k = 0,488
VARIABEL III. y = mx + c y = 0,032x + 1,932
m1 =
m2 =
m3 =
m4 =
Sehingga, k = 0,22
PERHITUNGAN CA MODEL DENGAN CA PERCOBAAN CA MODELVariabel I.
t = 0 menit..Ca = 0,1
t = 1 menit..Ca = 0,09
t = 2 menit..Ca = 0,083
t = 3 menit..Ca = 0,076
t = 4 menit..Ca = 0,071
Variabel II.
.
.
t = 0 menit..... Ca = 0,1
t = 1 menit.....,10Ca = 0,091
t = 2 menit.....,205Ca = 0,082
Variabel III.
.
.
t = 0 menit..... Ca = 0,1
t = 1 menit.....Ca = 0,093
t = 2 menit.....Ca = 0,087
CA PERCOBAANVariabel IPROSES KONTINYU
tVtitran (ml)Ca
01,80,036
11,60,032
21,50,030
31,40,028
41,50,030
Variabel IIPROSES KONTINYU
tVtitran (ml)Ca
01,50,030
11,50,030
21,40,028
31,40,028
Variabel IIIPROSES KONTINYU
tVtitran (ml)Ca
01,40,028
11,20,024
21,30,026
31,30,026
METODE RUNGE KUTTA Neraca massa totalinput output = akumulasi Fo - 0 = dV = Fo.dtV = Fo.t ............(1) Neraca massa komponen akumulasi = input output laju konsumsi konversi = Fo.Cao 0 V.k.Ca2
.........(2) Persamaan (1) dan (2) diselesaikan dengan orde 4k1 = k2 = k3 = k4 = Ca = Ca model = Cao + Cat = 1 menit Variabel 1 Etil Asetat 0,1 Nk = 1,012 Cao = 0,1tVHClCak1k2k3k4CaCa model
01,80,036-----0,1
11,60,032-0,067-0,013-0,0544-0,00377-0,03420,09
21,50,030-0,0053-0,0026-0,0032-0,002-0,003160,083
31,40,028-0,004-0,0025-0,0028-0,00195-0,002740,074
41,50,030-0,0033-0,0023-0,0025-0,00188-0,002470,071
Variabel 2 Etil Asetat 0,2 Nk = 0,448 Cao = 0,1
tVHClCak1k2k3k4CaCa model
01,50,030-----0,1
11,50,030-0,0113-0,0028-0,00568-0,00237-0,005080,091
21,40,028-0,0065-0,00321-0,00412-0,00248-0,003930,082
31,40,028-0,0053-0,00323-0,00412-0,00253-0,003620,082
Variabel 3 Etil Asetatk = 0,22 Cao = 0,1tVHClCak1k2k3k4CaCa model
01,40,028-----0,1
11,20,024-0,01139-0,00281-0,0062-0,00231-0,005280,093
21,30,026-0,00606-0,00305-0,00402-0,00235-0,003760,087
31,30,026-0,00522-0,00315-0,00375-0,00245-0,003580,087
