Analisis Data Absorbsi.docx

I.

Analisis Data 

Penambahan BaCl2 Tangki Umpan (S5) Waktu 0

: T2-T1 = 30.4 - 26.7 = 3.7 x 10/100 = 0.37 mL BaCl2

Waktu 10

: T2-T1 = 33.3 – 27.5 = 5.8 x 10/100 = 0.58 mL BaCl2

Waktu 20

: T2-T1 = 35.5 – 26.2 = 9.3 x 10/100 = 0.93 mL BaCl2

Waktu 30

: T2-T1 = 37.5 – 26.5 = 11 x 10/100 = 1.1 mL BaCl2

Waktu 40

: T2-T1 = 44 – 22.5 = 21.5 x 10/100 = 2.15 mL BaCl2

Tangki Umpan (S4)



Waktu 10

: T2-T1 = 45.8 – 17.3 = 28.5 x 10/100 = 2.85 mL BaCl2

Waktu 20

: T2-T1 = 46 – 10 = 36 x 10/100 = 3.6 mL BaCl2

Waktu 30

: T2-T1 = 46.5 – 5.3 = 41.2 x 10/100 = 4.12 mL BaCl2

Waktu 40

: T2-T1 = 50 – 3 = 47 x 10/100 = 4.7 mL BaCl2

Pembuatan Larutan NaOH 0,1 N 30 liter N=Mxi 0,1 = M x 1, M= 0,1 M M = (g/Mr) x (1000/V) 0,1 = (g/40) / (1000/30000) 120 gram = g (massa NaOH yang dibutuhkan)



Pembuatan Larutan Asam Oksalat Massa A.oksalat

= 0,633 gram

V

= 100 mL

BE

= Mr / 2 = 126 / 2 = 63 g/ml

N = (m/BE) x (1000/V) N = (0.633/126) x (1000/100) N = 0,1 N 

Standarisasi Larutan NaOH dengan Asam Oksalat V1 = volume larutan NaOH rata-rata titrasi = 25,55 mL V2 = volume larutan A.oksalat = 25 mL N1 = Normalitas larutan NaOH (yang dicari) N2 = Normalitas larutan asam oksalat = 0,1 N

N1 V1 = N2 V2 N1 x 25,55 = 25 x 0,1 N N1 = 0,098 N Saat t = 0 1. Inlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M 26,1

= (50 𝑚𝑙) × 0,1 M = 0,0522 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

30,4−26,1 50 𝑚𝑙

) × 0,1 M × 0,5

= 4,3 × 10−3 Outlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M =(

0 50 𝑚𝑙

) × 0,1 M

=0 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 0

= (50 𝑚𝑙) × 0,1 M × 0,5 =0 CO2 absorbed = laju alir × [(CN)o-(CN)i] = 0,033 × [0-4,3 × 10−3] = -1,419 × 10−4 1

= laju alir × 2 ×[(Cc)i-(Cc)o] 1

= 0,033 × 2 ×[0,022-0] = 3,63 × 10−4

Saat t = 10 2. Inlet 𝑇3

CNaOH (Cc)i = (50 𝑚𝑙) × 0,1 M 28,2

= (50 𝑚𝑙) × 0,1 M = 0,0564 𝑇2−𝑇3

CNa2CO3 (CN)i= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

33,3−28,2 50 𝑚𝑙

) × 0,1 M × 0,5

= 5,1 × 10−3 Outlet 𝑇3

CNaOH (Cc)i = (50 𝑚𝑙) × 0,1 M 11,6

= (50 𝑚𝑙) × 0,1 M = 0,0232 𝑇2−𝑇3

CNa2CO3 (CN)o= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

45,8−11,6 50 𝑚𝑙

) × 0,1 M × 0,5

= 0,0342 CO2 absorbed = laju alir × [(CN)o-(CN)i] = 0,033 × [0,0342-5,1 × 10−3] = 9,603 × 10−4 1

= laju alir × 2 ×[(Cc)i-(Cc)o] 1

= 0,033 × 2 ×[0,0564-0,0342] = 3,663 × 10−4

Saat t = 20 3. Inlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M

25,6

= (50 𝑚𝑙) × 0,1 M = 0,0512 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

35,4−25,6 50 𝑚𝑙

) × 0,1 M × 0,5

= 9,8 × 10−3 Outlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M 4,6

= (50 𝑚𝑙) × 0,1 M = 9,2 × 10−3 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 46−4,6

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 = 0,0414 CO2 absorbed = laju alir × [(CN)o-(CN)i] = 0,033 × [0,0414-9,8 × 10−3] = 1,0428 × 10−3 1

= laju alir × 2 ×[(Cc)i-(Cc)o] 1

= 0,033 × 2 ×[0,0512-9,2 × 10−3 ] = 6,93 × 10−4

Saat t = 30 4. Inlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M

23,5

= (50 𝑚𝑙) × 0,1 M = 0,047 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

37,5−23,5

) × 0,1 M × 0,5

50 𝑚𝑙

= 0,014 Outlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M 10

= (50 𝑚𝑙) × 0,1 M = 9,2 × 10−3 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 =(

46,5−10 50 𝑚𝑙

) × 0,1 M × 0,5

= 0,0365 CO2 absorbed = laju alir × [(CN)o-(CN)i] = 0,033 × [0,0365-0,014] = 7,425 × 10−4 1

= laju alir × 2 ×[(Cc)i-(Cc)o] 1

= 0,033 × 2 ×[0,047-0,02] = 4,455 × 10−4

Saat t = 40 5. Inlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M 18

= (50 𝑚𝑙) × 0,1 M = 0,036

CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 44−18

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 = 0,026 Outlet CNaOH

𝑇3

= (50 𝑚𝑙) × 0,1 M 14

= (50 𝑚𝑙) × 0,1 M = 0,028 CNa2CO3

𝑇2−𝑇3

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 50−14

= ( 50 𝑚𝑙 ) × 0,1 M × 0,5 = 0,036 CO2 absorbed = laju alir × [(CN)o-(CN)i] = 0,033 × [0,036-0,026] = 3,3 × 10−4 1

= laju alir × 2 ×[(Cc)i-(Cc)o] 1

= 0,033 × 2 ×[0,036-0,028] = 1,32 × 10−4