/ vilim: i(ym hns Ani yvWị> rirHij nhj' r.'H
dich NaOH IM thi se thu diTdc muoi gi ? Co khoiliTdng bao nhidu ?
Gidi ^lru,(v:.:: :'•
D$t CTTQ cua hai olefin : C-H^-. " ' ' ' ' ' ' a) Kh^i lUdng binh tSng 20,16g d o chinh la khoi liTdng ciia hai olefin. ' * ' a) Kh^i lUdng binh tSng 20,16g d o chinh la khoi liTdng ciia hai olefin. ' * '
m c - u - = 20.16(g) Z > M A = ^ = 50.4 => 14n = 50,4 => n = 3,6 => ni = 3 < n = 3,6 < n2 = 4 , j , ^ , . , , , _ => 14n = 50,4 => n = 3,6 => ni = 3 < n = 3,6 < n2 = 4 , j , ^ , . , , , _ C T P T hai olefin: * Cdc d6ng phan olefin : , U CjHfi: CH3-CH=CH2 propen ^ S U ' " O J ^ - ^ > xS 5 U C 4 H X : CH3-CH2-CH=CH2 but-1-en * ' Y ' ' ^ CH3-C=CH2 2-metylpropen i i A 1 • : i , r ; < r t ' u : ' . i , h i CH3-CH=CH-CH3 but-2-en , : O c f , ' > £ T ' ^ ; ^ H CH H3C^'^=^"CH3 H 3 C ^ ^ ^ H " ' cis-but-2-en trans-but-2-en
Phan djing va phuong phap giai H6a hqc 11 HOu CO - D8 Xuan HUng b) mol NaOH : HN^OH = 1 x 0,25 = 0,25 (mol)
CjHs + ^Oj >- O 2 > 3CO2 + 3H2O 3 C 0 2 + 3 H 2 O -.l^'i" :0;T' ịÓ 'tOj';
x m o l 3 x m o l
C4HX + 6O2 > 4CO2 + 4H2O
y mol 4y mol -.'ijrr •,^ , \,L^»|>O* - . H V ; : - ^ ' ' ' " ' "
3 x+4v fx = 0,16
Ta C O : ^ = 3,6 <=> x + y = 0,4 =><^ (mol) oAi
x + y [y = 0,24 v^--;--y,:v ^^,^^01.^:
n c o j = 3 x + 4y = 1,44 • • , .jrv-jiir
Ta C O ti 10 : "'^""^ = — = 0,17 < 1 => thu difdc muoi axit NaHCOj. %
"CO2 1^44 ^ ^
,„;. CO2 + NaOH > NaHCOs ,t 1^ mil, . i < 1 ' (f n fi '' i
in 0,25 mol 0,25 mol , IK,-« >u not 1 ' ^A^Hl
Khoi liTdng muoi : m = 0,25 x 84 = 21 (g).
Bai 10. Dot chay mot hidrocacbon A thu diTcfc so mol CO2 bkng 2 mol H2Ọ Cho 0,05 mol A phan iJng vdi dung dich AgNOa/NHj diT thu difdc 7,95g ket tuạ Xac djnh cong IhuTc cau tao cua A va hôn thiinh sd do phan itng :
C _ A l 2 % . X ^ " 2 ° ) y _ L 5 0 0 ! c ^ z ^ A B > Cao su c6 clọ
1 : 1
D a i C T T Q c u a A : Q H , . > , , .»M M'-H.) , 1 HO
x + y
4 O2 > x C 0 2 + ^ H 2 0 , , , ,wi ' , r)'M4'!?,miM.iBm
Thco do : n c o j = 2nH20 => x = 2.-^ => x = y • '
Q H y + AgNO,, + N H j >.C,Hỵ,Ag + NH4NO3 ' ' ^
0,05 mol 0,05 mol ''
(VI M i = — = 159 => trong A chi cd I H hnh dong)
0.05 .r'fMir
T a c d : 12x + y + 107 = 159 o l 2 x + y = 52 i t i - , ' l . '
ma x = y =>x = y = 4 / ,M' ( ' / l ' »>lí" i '
Vay CTPT A la C4H4. CTCT : CH2=CH-C=CH. , » .) •{} Phifdng trinh phan ilng theo sd do : , ( (adsbygoogle = window.adsbygoogle || []).push({});
9C + 2AI2O3 AI4C3 + 6 C 0 ,• (,(,. a > '"; >\rf
. , ' (X) f f , }?
AI4C3 + I2H2O > 4Al(OH)3 + 3CH4t , ;^ ( Y ) 1 2 0 2CH4 -i^02!£^ C2H2 + 3H2 2CH^CH ^ CH2=CH-C=CH CH2=CH-C=CH + HCl CH2=CH-C=CH2 ^ 'r'^' • (B) CI nCH2=CH-C=CH2 -(-CH2=CH=C-CH2 (cao su C O clo).
B a i 11. Dot chay 560cm'' hon hdp khi (dktc) gom 2 hidrocacbon cd cdng so nguyer turcacbontathuduWc4,4gC02va l,9125ghdiniA3c. ) ; A i , ,1;; • ,
a) X d c d i n h C T P T c a c c h a t h t r u c d . :jfa,,,.^,.y!. ' , .(isrsp :H*' , . | , } H . , ' - '^'^
b) Tinh ^okhoi liTdng cac chát.
c) Neu cho luTdng CO2 tren vao 100 ml đ K O H 1,3M. Tinh C M muoi tao thanh.
Gidi
O b^i nay, ta dung phifdng phap so nguyen tuf H trung binh ket hdp vdi phUdng phdp bi^n luan de giaị
a) Xac dinh CTPT cac hidrocacbon : Dat CTPT 2 hidrocacbon tren : A:C,Hy
B: C^Hỵ CTPT trung blnh 2 hidrocacbon tren : C - H -
A y
,1 , ! { 1 < s i .
Gia suf y < y ' y < y < y '
0.56 14'
,<0 I1
So mol hon hdp khi nhh = —— = 0,025 mol 22,4 nco2 = 4,4/44 = 0,1 (mol) • ^ ' n U , j , h i ^ 1 " 1 nHjO = 1.9125/18 = 0,10625 (mol) ' ' ^ u « t 0,025 x + y 4 . O 2 — ^ x C 0 2 + ^ H 2 0 0,025 X 0,025 y/2 x = 4 "CO2 =0,025x = 0,1 nHjO = 0 . 0 2 5 ^ = 0,10625 [y = 8,5 , , ' ^ CTPT A, B cd dang : A : C4Hy va B : C4Hỵ ' ' , ' ' ' Ta cd y < y < y ' hay y < 8,5 < y ' (1) - j ; ; ^ 121
Phin dgng va phuang phap giai H6a hpc 11 HOu oa-06 XuSn Hunp
Bi^n lu|ln tim C T P T B : 8,5 < ý chSn
Lý < 2x + 2 = 2.4 + 2 = 10 =>y'=10=>CTPTB:C4H,o =>y'=10=>CTPTB:C4H,o TiTdng W bi$n lu|in tlm C T P T A : y < 8,5 2 4 8 A C 4 H 2 C 4 H 4 C 4 H 6 C 4 H K Vay c6 4 cSp nghi^m : fA:C4H2 A:C4H4 va • A:C4H6 B: C4H|Q' B: C 4 H B: C4H,().
c) Tinh C M cdc muoi tao thanh : n K O H = V . C M = 0,1.1,3 = 0,13 (mol) n K O H = V . C M = 0,1.1,3 = 0,13 (mol)
' A :C4 H 8 .'H^ SJ n - K ; : j . » . ^ i i
B:C4H,o- r3?1^tj5:jHX •
I i>Oi fu;v rfirit j^ápSJi 0(:J:i oat'I {
Ta cd : JlKOH = ^ = 1,3 =^ Tao th^nh 2 muoị nco2 0.1
CO2 + 2 K 0 H -> K2CO3 + H2O
a 2a a (mol)
C O 2 + K O H -> K H C O 3
b b b (mol)
a + b = nQQ^ = 0,1 (adsbygoogle = window.adsbygoogle || []).push({});
)Jfv t[ i i r t' i ,1)11' <t;.iKj
Ta c(5 : a = 0,03 'ij&r^yjtisrt rinfd f c o « T ' t T 3 (mol) 2a + b = tiKOH = 0.13 [b = 0,07
C M ( K 2 C 0 3 ) = • ^ Y = 0 , 3 ( M ) ; C M ( K H C O 3 ) = ^ Y = = 0 , 7 ( M ) ,s?,filofn6?
Bai 12. Dot chay hôn toan hon hdp gom ankin (A) va ankan (B) c6 V = 5,6 lit
(dktc) di/dc 30,8g CO2 va 1 l,7g H2Ọ
Xdc dinh C T P T A,B. Tinh % A,B. Biet B nhieu hdn A mOt C
Gidi ' ^ • •
6 bai nay, dot chay hon hdp 2 hidrocacbon khong phai la dong dSng cua nhau nen khong diing phÚdng phip trung binh dÚdc m^ suT dung phi/dng phip ghdp I n