Gaz chiqishi uchun gomlog tarkibida uglerod atomlari 7 tadan kòp bòlishi kerak yani Benzoy k-ta da 7 C mavjud

C6H5COOH --- > 6 ta H

50 % --------- 6
100 % ----- x = 12ta H dastlab arenda

CnH2n-6

2n - 6 = 12
2n = 18
n=9

C9H12 ----- > C6H5COOH + 2CO2
122 -------------- 44.8
x=61✅ ---------- 22.4

#A

42.14 / 6.02=7 ta H atomi

R-COOH + NaOH = RCOONa+H2O
1 ------- 1
0.2=x-------- 0.2mol

R-COOH+NaHCO3 = CO2
1 --------------------- 1
0.5=x ----------------------- 0.5

Jami 0.5+0.2=0.7mol

CnH2nO2
1mol -------------- 2n
0.7mol----------- 7

7=1.4n
n=5 ---- > C5H10O2 ( valerian k - ta )

C4H9COOH+Ba = (C4H9COO)2Ba

3390.7=237.3 g ✅*

#A

16 / 2 = 8

Demak 8 ta elektron bergan

CH4 --- > C - 4

- 4 - 4 = + 4

CCl4 --- > C + 4

CCl4 = 154 ✅

35.2 - 29.8 = 5.4 H2O

5.4 / 18 = 0.3

x ------------------ x --------- x
C5H11OH ---- > C5H10 + H2O

0.4mol ------‐---------------- 0.3 mol

2C5H11OH --- > C10H22O + H2O
2y ----------------- y --------------- y

35.2-29.8=5.4
5.4/18=0.3

x+2y=0.4
x+y=0.3
y=0.1
x=0.2

C10H22O = 158 * 0.1 = 15.8 ✅

532ni xam korib berilar

Orqadan ketamiz

Nima uchun 22 g ga kamaydi?

Karbon k-tani Na li tuzlari natron ohak qoshilganda alkan chiqadi.

Alkan gaz shu hisobiga kamaydi

Ma'lumot òrnida alkanlarni dastlabki 4 vakili gaz. Demak pentan va undan keyingi gomlogi tòĝri kelmaydi.

R-COONa + NaOH = R - H + Na2CO3
0.5 ------------------- 22
1 --------------- x = 44

CnH2n+2 = 44

14n + 2 = 44
n = 3 ---- > C3H8 ( propan 3 - vakil)

Demak murakkab efirni k-ta qismi C4H8O2

Sabab - n = 3 ( C3H8 )
bitta C atomi Na2CO3 tarkibidaga òtb ketdi, u karbon kislotani funksional guruhida edi

3 + 1 = 4 -- > C4H8O2 ( moy k - ta )

Dastlabki jumlalarga qaytamiz.

Murakkab efir gidrolizlanganda
CH3OH chiqqan ekan, demak spirt qismi CH3^-

Demak umumiy efir

K-ta qism 4 + spirt qism 1 = 5

C3H6COOCH3 -- > moy kislotani metil efiri yoki metil butanoat

C5H10O2 umumiy for-a Mr = 102

Bizdan massa soralgan

C5H10O2 ---- > CH3OH
102 --‐-----‐------ 1 mol
x=51✅ ---------- 0.5 mol

#D