sunfat, chiing to phan ling xay ra vira diị
- Tir ti le mol: n : nH ==2:5
San phim khii la SÔ (khdng th^ la hoac S).
Phucfng trinh phan urng:
2Fe + 6 H 2 S O 4 FẹCSÔ), + SSOj + 6H2O (1)
Theo (1): S6' mol electron (e) sat cho = 2 s6' mol SO2
= s6' mol H 2 S O 4 = ỵ
2.16. Dap an dung la Ạ 2.17. Dap an dung la D. 2.17. Dap an dung la D.
Khi giam ap suat ciia h6, can bang (IV) chuy^n dich theo chi^u nghich.
2.18. Dap an diing la B.
AH > 0: phan ling thu nhifit nSn khi tang nhiet d6 can bang chuy^n dich theo chilu thuan. dich theo chilu thuan.
2.19. Dap an diing la Ạ
Phuong trinh phan ling:
5Na2SO, + 2KMn04 + 6NaHS04 8Na2SG4 + 2MnS04 + K 2 S O 4 + 3H2O
T6ng he s6' cua cAc chat = 5 + 2 + 6 + 8 + 2 + 1+ 3 = 27.
2.20. Dap an dung la D.
o - 2 + 4
4S + 6NaOHdac-).2Na2S +Na2S03 + SHjO 3. Sl/DIEN LI 3. Sl/DIEN LI
3.1. Dap an dung la Ạ
Phuong trinh phan ting trung hoa :
M(OH)2+H2S04 -)-MS04 + 2H20 (1)
Gia sir c6 1 mol (98 gam) H2SO4 phan ling. Khi do kh6'i luong dung Q8 inn Q8 inn
dich axit H2SO4 la = 490 gam.
Theo (1): Sau khi hoa tan 1 mol-> S6'mdungdich mol MSO4 = 490 + M + 34 tao thanh la : (490+M + 34).27,21 ^ ^ _^ ^ 64 M(OH100(M + 96) )2, kh6'i lugng dung dich \k : v a y kirn loai M la Cụ v a y kirn loai M la Cụ
3.2. Dap an diing la Ạ
Chi CO nhung mu6'i tao b6i baza manh va axit yéu, khi thuy phan m6i
cho pH > 7.
3.3. Dap an dung la D.
Phuang trinh phan ling : KOH + HCI ^ KCl + HjO S6' mol KOH = 0,1 . 1 = 0,1 (mol). S6' mol KOH = 0,1 . 1 = 0,1 (mol).
Néu KOH phan ling h6t thi chát tan la KCl phai c6 kh6'i lugng
0,1.74,5 = 7,45 gam 16n hem 6,525 gam.
Nhu vay KOH du, HCI phan ling hét. Chat tan g6m KCl va KOH dụ
Ggi s6' mol HCI phan ling la x s6' mol KCl la x. Ta c6 : (0,1 - x).56 + 74,5x = 6,525 -> x = 0,05 ^ CM(HCI) = ^ = 0.5M. (0,1 - x).56 + 74,5x = 6,525 -> x = 0,05 ^ CM(HCI) = ^ = 0.5M.
3.4. Dap an diing la Ạ
Dung dich trung hoa difin n6n :
S6' mol dien tich duang = s6' mol dien tfch am
-> 0,2.2 + 0,3 = x + 2y x + 2y = 0,07 (I) ^ 0,02.64 + 0,03.39 + 35,5x + 96y = 5,435 gam (II) ^ 0,02.64 + 0,03.39 + 35,5x + 96y = 5,435 gam (II) P Giai he phuang trinh (I) va (II) dugc x =0,03 ; y = 0,02.
3.5. Dap an dung la Ạ
Metyl amin va amoniac d^u c6 tinh baza nSn lam xanh quy tim. Natri axetat la mu6'i cua baza manh va axit yS'u, trong dung dich bi Natri axetat la mu6'i cua baza manh va axit yS'u, trong dung dich bi thiiy phan cho m6i trudng baza nSn cung lam xanh quy tim.
3.6. Dap an dung la D.
Trong dung dich :
- Axit HCI d i e n li hokn toan : HCI -> H+ + Cl"
- Axit C H 3 C O O H d i e n li 1% : CH3COOH < > H^ + CHjCOO". Dung dich HQ c6 pH = x -> [ H+ ] = CM(HCI) = 10"'' = CM( C H 3 C O O H ) . Dung dich HQ c6 pH = x -> [ H+ ] = CM(HCI) = 10"'' = CM( C H 3 C O O H ) .
Dung dich C H 3 C O O H c6 [H+ ] = 10"" . 10"^ -> pH = y = x + 2.
.7. Dap an diing la Ạ
Dat V = 1000 ml. Ta c6:
Sau khi thuc hien phan ling trung hoa: H* + OH" HjO S6'mol = 0,02 mol/2 lit [H^] =0,01M pH = 2. S6'mol = 0,02 mol/2 lit [H^] =0,01M pH = 2.
3.8. Dap an dung Ik B.
4 chat dien li la:
KA1(S04)2.12H2O, CH:,COOH , CăOH)2 va CH^COONfH 4
3.9. Dap s6' diing la D.
pH = 1 ^ [H^] = 0,1 mol// -> n^. = 0,1.0,1 = 0,01 mol pH = 12 -> [ H 1 = 10 ^ [ 0 H 1 = 10-2 pH = 12 -> [ H 1 = 10 ^ [ 0 H 1 = 10-2
-> n^H^ =0,01.0,2 = 0,002 (mol).
S6' mol OH" CO trong 100 ml dung djch NaOH a mol// la: 0,1.a mol. ^ 0,l.a = 0,01 + 0,002 = 0,012. ^ 0,l.a = 0,01 + 0,002 = 0,012. 0,012 ,,, a = — — =0,12 mol//. 0,1 3.10. Dap an diing la B. 3.11. Dap an diing la C. S6' mol cac chát:
nNaOH " 0,25.1,04 - 0,26 (mol); np,ci, - 0,024 mol; °H,so, =0,04 mol; n^, (SQ,), =0,016 mol °H,so, =0,04 mol; n^, (SQ,), =0,016 mol
Cac phirong trinh phan ling:
2NaOH + H2SO4 > Na2S04 + 2H2O
0,08 mol 0,04 mol
3NaOH + FeCl3 >Fe(0H)3 i + 3NaCl 0,072 mol 0,024 mol 0,024 mol 0,072 mol 0,024 mol 0,024 mol
6NaOH + Al2 (804)3 >2A1(0H)31+ 3Na2S04
0,096 mol 0,016 mol 0,032 mol
Sau ba phan ling (1) (2) va (3), s6' mol NaOH con du \h:
"NaOH =0,26-(0,08+ 0,072+ 0,096) = 0,012 (mol) NaOHdu + A1(0H)3 i > Na[Al(0H)4 ] NaOHdu + A1(0H)3 i > Na[Al(0H)4 ]
0,012mol 0,012mol
Ke't tua g6m: 0,024 mol Fe(0H)3 va 0,032 - 0,012 = 0,02 (mol) Al(OH)3. -> m = 0,024.107 + 0,02.78 = 4,128 (gam). -> m = 0,024.107 + 0,02.78 = 4,128 (gam).
3.12. Dap dn dung la C.
(1) (2) (2) (3)
J.13 Dap an diing la B. S6' mol trong 100 ml dung dich h6n hop hai axit:
n^. = 2.0,1.0,05 + 0,1.0,1 = 0,02 (mol)
S6' mol OH" trong 100 ml dung dich h6n hop bazo: n^H- = 0,1.0,2 + 2.0,1.0,1 = 0,04 (mol). n^H- = 0,1.0,2 + 2.0,1.0,1 = 0,04 (mol). Khi tr6n iSn hai h6n hop dung dich xay ra phan iJng:
H^ + OH -> H2O
-> Trong 200 ml dung dich c6 0,04 - 0,02 = 0,02 mol OH". Trong 1000 ml dung dich c6 a mol OH . Trong 1000 ml dung dich c6 a mol OH .
1000.0,02 ^ , ,^ = 0,1 mol hay 10 = 0,1 mol hay 10 a = H^ 200 10 , - 1 4 10 , - 1 4 OH- IO = 10 , - 1 3 3.14, 3.15, - ^ p H = 1 3 . Dap an diing la B.
Cac phuong trinh phan ling:
K H C O 3 + HCl KCl + COjt + H2O
Na2C03 + 2HC1 ^ 2NaCl + C02t + H2O
NajCO, + CO2 + H2O 2NaHC03
Theo (1) : ncô(,) = DKHCO, =0,lmol = nHci(i)
Theo (2): nco^(2) ^\^HOi2) =^(0,2-0,l) = 0,05(mol).
S6 mol NajCO, du sau phan ling (2) 1^:
"Na,co,du =0.15-0,05 = 0,1 mol. Theo (3): nco^(3) = nNa^co,du = 0,1 mol. Theo (3): nco^(3) = nNa^co,du = 0,1 mol.
-> s6' mol CO2 bay ra la: 0,1 + 0,05 - 0,1 = 0,05 (mol). Vay Vco, = 0,05.22,4 = 1,12 lit. Vay Vco, = 0,05.22,4 = 1,12 lit.
Dap an diing la B. S6' mol cac chát: S6' mol cac chát:
3,808
"co, = 22,4 = 0,17 (mol); "H,O ~ ~rr = 0,3(mol). 5 4 18
Hhuang trinh phan iJng chdy:
C„H2„,20fy02->C02+(n + l)H20 (1)
X mol 0,17 mol 3n 3n
— - -> XX 0,17 = 0,255 mol. Theo dinh luat bao toan kh6'i lugng: Theo dinh luat bao toan kh6'i lugng:
-> m,„,„| = mcô + m H p -mô = 7,48 + 5,4-8,16 = 4,72 gam. = 7,48 + 5,4-8,16 = 4,72 gam.
3.16. Dap an dung la C.