- 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
3. Go ip la s6'mol ciia C.H.nÔ trong X, ta c6 phuong trinh v^ tdng s6' mol cacbon hoac CỘ
mol cacbon hoac CỘ
6p + 5(0,18-p) = 1,05. Rut rap = 0,15 mol. Rut rap = 0,15 mol.
%C,H.COOC,H,=5llM!1100%. 85,07%.
%CH3COOC3H, = 100% - 85,07% = 14,93%. 4. Ggi R la g6'c hidrocacbon trung binh cua 2 axit, ta c6: 4. Ggi R la g6'c hidrocacbon trung binh cua 2 axit, ta c6:
R -COOR + NaOH R -COONa + ROH (2)
S6 mol mu6'i = s6' mol este = ^'^^"^'^^^ = 0,027 (mol).
Do do M„ 2,529 0,027 0,027
20,1
= 93,66 R = 93,66-67 = 26,66 .
G6'c R < 26,66 chi c6 thé la C H , - (M = 15) va g6'c R > 26,66 la CjH.,-(M = 29). Do do c6ng thirc cua cac este va mu6'i tuang iJng: (M = 29). Do do c6ng thirc cua cac este va mu6'i tuang iJng:
CH3-COOC3H, va CH3-C00Na (M = 82). QH,-COOC3H5 va C^Hj-COONa (M = 96). QH,-COOC3H5 va C^Hj-COONa (M = 96).
V l s6' mol mud'i cung tl 16 nhu s6 mol este nfin s6' mol CjH^-COONa = 0,027.^^ = 0,0225 (mol) va khd'i luang Q H s C O O N a = 0,0225.96 = 2,16
0,18
(gam) va khoi luang CHj-COONa = 2,529 - 2,16 = 0,369 (gam).
Thi du 2. Cho mOt h6n hop long g6m ancol etylic va hai hidrocacbon
la d6ng dang lien tiep. Chia hOn hop thanh 2 phSn bang nhaụ Lam bay hoi
hét phan 1, thé tich hoi ciia no bang thi tich ciia 1,32 gam khi cacbonic dupe do d cung 6iiu kien nhiet dO, ap suát.
Di dot chay het phan 2 cdn dung lupng Oj thu dupe bang each nhiet phan hoan toan 92,43 gam KMn04. Cho san phdm chay lin lupt di qua binh 1 dung
H2SO4 dac va binh 2 dung dung dich Bă0H)2 (du). Sau thi nghiem khoi lupng binh 1 tang 3,915 gam, cf binh 2 c6 36,9375 gam két tiia trang.
Hay xac djnh c6ng thiie phan tir cua hai hidrocacbon va tinh thanh
ph^n % theo khO'i lupng ciia mOi chat trong h6n hpp Ạ
Hu&ng đn gidi:
Theo phan itng phan huy:
2KMn04 — ^ K2Mn04 + Mn02 + Ô^
va theo sir háp thu H2O cua H2SO4 va háp thu CO2 ciia BăOH)2:
Bă0H)2 + CO2 BaCO,i + HjO 92,43 158.2 3,915 18 36,9375 „ ncc), = n„.co, = - 0-1875 (mol).
Gpi c6ng thirc ciia 2 hidrocacbon la C,H^ (trong d6 x,y la SNTCTB va SNTHTB) va a, b la s6' mol ciia C 2 H, 0 H va cua 2 hidrocacbon, theo di bai
1 32
a + b = - — = 0,03 (mol). Theo cac phan ling chay:
44
C2H,0H + 3O2 -> 2CO2 + 3H2O (3)
Ta c6: n„, = = 0,2925 (mol).
n H, o= ^ 7: r - = " 0.2175 (mol).
O j ^ x C O j + ^ H j O (4) 4y 2
Ta tháy s6' mol ciia CjH^OH = so mol nguyen ttr oxi c6 trong ancol = tdng s6' mol O trong CO2 va trong H2O trir di so mol O dem dÓt chay, nen:
a = (0,1875.2 + 0,2175 - 0,2925.2) = 0,0075 (mol). va b = 0,03 - 0,0075 = 0,0225 (mol). f „ _ 0,1875-2.0,0075 _ , . Dodo: x = = 7,66. 0,0225 _ (0,2175-3.0,0075) ,^ , y = = 17,3. 0,0225
Vay CTFT cua cac hidrocacbon la CjHif, ( M = 100) va CsH,x ( M = 114). Gpi b, la s6' mol cua C^U,^, ta c6 phuong trinh v^ tdng s6' mol CO2:
8b, + 7(0,0225 - b,) + 2.0,0075 = 0,1875 ( Q H „ (QH„) (C2H,OH)
Rut ra: b, = 0,015 mol va s6'mol Cjlii^, = 0,0075 mol.
KhÓi lupng cac chat:
mc,„,oN = 0,0075.46 = 0,345 (gam).
mc,H„ = 0,015.114= 1,71 (gam). m^^i,^^ = 0,0075.100 = 0,75 (gam).
=> %C2H,0H = 12,3%; %C,H,x = 60,96%; %CyH,^ = 26,74%. V I I I . X A C D I N H CONG THtTC PHAN TU"
CUA CAC HOP C H A T V 6 CO
Xac djnh cOng thiJc phan til ciia cac hpp chát v6 ca eo nhi^u each khac nhaụ Sau day xin neu m6t s6' each thOng dung va don gian nhát.
A L C a c h t h u r n h a t
Xac dinh CTFT bang each lap ti le so nguyen tu cua cac nguyen to
|trong phan tii nhu xac dinh cOng thiic don gian (CTDG) trong hoa hffu cạ |.Ca« luu y rang, doi vdri cac hpp chát v6 ca, tuyet dai da s6' trucfng hpp
'•CTDG cung dong tlidi la CTPT.
Thi du Ị De hoa tan 4 gam oxit Fe.Oy c^n 52,14 ml dung dich HCl 10% |<D = 1,05 g/ml). Xac dinh c6ng thiic phan tir cua sat oxit.
Hu&ng đn gidi:
Phuong trinh hoa hoc ciia phan ling hoa tan sat oxit:
Fe.Oy + 2yHCl -> xFeClj^/, + yH20 (1)
(Hoc sinh thudng lung tiing \i hoa tri ciia Fe: CSn chii y t6ng hoa tri