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  LEMBAR PERHITUNGAN REAGEN 1.    NaOH 0,185 N =1000    0,185 =40 /1000100 98100  W = 0,755 gram 2.   Volume katalis H 2 SO 4  0,185 N =  −     =67,58 −22,7524,77=1,81 /   =1000   98100   0,185 =1,81   .98 /1000285  2  98100   =1,45   3.   Volume katalis HCl 0,185 N =  −     =51,0125 −22,7524,77=1,141 /   =1000   98100   0,185 =1,141   .36,5 /1000285  2  98100   =6,74   4.   Kebutuhan Minyak Jarak dan air a.   Volume basis = V minyak jarak   + V aquades  + V emulsifier   + V katalis  Volume basis - V emulsifier - V katalis = V minyak jarak + V aquades Untuk katalis HCl : (285  –   15  –   6,74) ml = V minyak jarak + V aquades 263,26 ml = V minyak jarak + V aquades . ................1)   Untuk katalis H 2 SO 4  : 285  –   15  –   1,45 = V minyak jarak + V aquades 268,55 ml = V minyak jarak + V aquades . ................2)  b.   Perbandingan mol minyak jarak dengan air   =  −       =46,15 −22,7524,77=0,961 /   110=        110= ,    /,    /  0,0321 V minyak jarak = 0,05507 Vaquades  = 0,05507 0,0321     =1,715 . ................3) Substritusi persamaan 3 ke dalam persamaan 1 (untuk katalis HCl) 263,26 = V minyak jarak + V aquades 263,26 = 1,175 V aquades  + V aquades  V aquades  = 97,64 mL V miyak jarak = 263,26  –   97,64 = 165,62 mL Substritusi persamaan 3 ke dalam persamaan 2 (untuk katalis H 2 SO 4 ) 268,55 = V minyak jarak + V aquades 268,55 = 1,175 V aquades  + V aquades  V aquades  = 98,65 mL V miyak jarak = 268,55  –   98,65 = 169,9 mL  LEMBAR PERHITUNGAN 1.   Perhitungan Kadar Asam Lemak Bebas (ALB) dalam Bahan Baku %=(.)          1000 100%   %=(3,5   0,185 ) 1067 /10   0,96484   1000  100%   %=7,16 %  %trigliserida = 100% - %ALB %trigliserida = 100% - 7,16 % = 92,839 % 2.   Perhitungan Konsentrasi Asam Lemak Bebas (ALB)  =(.)         =3,5   0,185 10   0,961 /=0,067  ⁄   =%%     = 92,839 %7,16 % 0,067 =0,868 /  3.   Analisa Asam Lemak Bebas yang Terhidrolisa  =(.)            = ℎ−   Ca bereaksi =            =        Pada pemberian katalis HCl 0  ℎ= 20   0,185 10   0,961 /= 0,385        = 0,385 −   0,067 =0,318       =  (0,318 )0,868 =0,122      t 5 menit=22   0,185 10   0,961 /=0,423        = 0,423 −   0,067 =0,356       =  (0,356 )0,868 =0,136   t 10 menit=24   0,185 10   0,961 /=0,462        = 0,462 −   0,067 =0,395       =  (0,395 )0,868 =0,152   t 15 menit=28,5   0,185 10   0,961 /=0,548        = 0,548 −   0,067 =0,481       =  (0,481 )0,868 =0,185   t 20 menit=31   0,185 10   0,961 /=0,597        = 0,597 −   0,067 =0,53      =  (0,53 )0,868 = 0,2   t 25 menit=34   0,185 10   0,961 /=0,655        = 0,655 −   0,067 =0,588   
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