- Dieii die kim logi kiem: Dien phan mud'i clorua hoac hidroxit nong chaỵ Dien che kim logi kiem tho: Dien phan mu6'i clorua nong chaỵ
a)Hai hop chat Id dong đng lien tiep (C„ vd C„ +,)
Day la trudng hap dan gian nhat ap dung phuang phap SNTCTB, v i n < n < n + 1, nghTa la neu t i m ducfc ii = 2,7 chang han t h i mot chat c6 s6 nguyen t i i C la 2, chat kia la 3.
Thi du Ị Dot chay 3,075 gam h6n hgp 2 ancol đng dang cilia ancol metylic va cho san pham Ian lugt di qua binh m6t dung H2SO4 dac va binh hai dung K O H ran. Tinh khoi lugng cac binh tang len, biét rang néu cho lugng ancol tren tac dung v6i natri tháy bay ra 0,672 1ft hidro (cf dktc).
Huofng đn gidi:
Goi n la s6' nguyen tiir cacbon trung bmh trong hai ancol, ta c6 cac phan irng:
C n H 2„ . , O H +y O. - ^ n C O j + Cn+DHjO (1)
H2SO4 dac hiit nuac, K O H ran háp thu CO2:
2 K 0 H + CO2 ^ K2CO3 + H2O (2)
2C„H2„„OH + 2Na ^ 2QH2„„ONa + H ^ t (3)
T6ng so m o l ctia hai ancol, theo (3) bang: x = 2 . - ^ ^ - = 0 , 0 6 (mol).
Vay K L P T T B ciia 2 ancol bang: ^ = 5 1 , 2 5 = 1 4 n . 18 -0,06
Tit do ta c6:
14 Theo (1) ta c6:
Kh6'i lugng binh 1 tang = khoi lugng nude =
= x(n + 1)18 = 0,06 (2,375 + 1).18 = 3,645 (gam). Khg'i lugng binh 2 tang = khoi lugng CO2 =
= x.n.44 = 0,06.2,375.44 = 6,27 (gam).
Thi du 2. H6n hgp X g6m 0,01 mol natri fomat va a m o l 2 mu6'i natri
cLia 2 axit no, dan chiJc la dong dang lien tiep. D6't chay h6n hgp X va cho san pham chay (COj, hai H2O) Mn lugt di qua binh 1 dung H2SO4 dac va binh 2 dung K O H ran tháy kh6'i lugng binh 2 tang nhieu lion binh 1 la 3,51 gam. PhSn chat ran Y con lai sau khi dot la NajCO,, can nang 2,65 gam.
1. Xac dinh c6ng thirc phan ti^ va ggi ten hai mu6'ị 2. Tinh % khoi lugng m 6 i mudi trong h6n hgp X .
Ha&ng đn gidi:
1. Ggi c6ng thiic chung cua 2 mudi chua biét la C^Hji^^iCOONa (trong do n la sd nguyen tur C trung binh ciia hai gdc axit) va tdng sd m o l la ạ Cac phan irng chay:
2HCOONa + O2 - > NajCO, + CO2 + H2O (1)
2C-H2,„COONa + (3n + 1 ) 0 , ^ NạCO, + (2ii + l)CO, + (2n + DH^O (2) Theo ( 1 , 2 ) khoi lugng ciia Y (NajCO,) bang:
a ^ 0 , 0 1
U 2 ,
va hieu khd'i lugng CO2 va H2O la:
.106 = 2,65 => a = 0,04 mol.
,,0,04 0,01
(2n + l ) + 4 4 - ( 2 n. , ) M l , M l 18 = 3,51
2 2
Rut ra n = 2,75. Vay cdng thiic cua cac mudi la: CjHs-COONa va C^H^-COONạ
natri propionat natri butirat hoac natri isobutirat (adsbygoogle = window.adsbygoogle || []).push({});
2. Ggi x la sd mol C^H^-COONa thi sd mol CzH^-COONa la (0,04 - x), ""eo ri ta c6:
_ 3x + 2 ( 0 , 0 4 - x ) n =
Vay: %HCOONa = 001^68^00% ^ 0.68-100% ^ 0,01.68 + 0,01.96 + 0,03.110 4,94
%QH,-COQNa="'"^f^-^°"^-^19,4%.
%C,H,-COONa = 100% - 13,8% - 19,4% = 66,8%. va s6' mol QH^OH = 0,03.5 - 0.01 = 0,025 (mol). va s6' mol QH^OH = 0,03.5 - 0.01 = 0,025 (mol).
0,025.60.100%
,, OH "^--^ = 66,96%.
% m c^ „ , „ H = 100 - 66,96 = 33,04%.
b) PhUffng phdp dUng so nguyen tu: cacbon trung binh (SNTCTB) vd so nguyen ti hidro truifg binh (SNTHTB)