C. Zn D Mg Ạ Al B Bạ
50% B 75% C 80% D 70%.
Hit&ng đn gidi
Chpn mx = 100 gam -> m c a c o ^ 80 gam va khoi li/gfng tap chat bkng 20
gam.
CaCOs CaO + CO2 (hieu suat = h) Phufdng t r i n h : 100 gam 56 gam 44 gam
™ , . «^ 56.80 , 44.80 , P h a n u n g : 80 gam ^ "Too"' ~100~ K h o i li/ong chat r ^ n con l a i sau k h i nung la
4 4 . 8 0 . h n^x - nico, = 100 - 100 56.80 ^ 45,65 -.n = 1 0 0 - 44.80.h 100 100 100
=> h = 0,75 -> hieu suat p h a n ufng bang 75%. Chon dap an B.
B a i 11: Cho h6n hop gom ba muoi MgClz, NaBr, K I vdi so mol ti/cfng ufng la 0,2 mol; 0,4 mol va 0,2 mol. Hoa t a n hon hop A t r e n vao nUdc tao ra dung dich X. DSn V(/) CI2 sue vao dung dich X, c6 c a n dung dich sau p h a n iJng thu diTpc 66,2g c h a t r a n . T i n h V (dktc).
Ạ 2,241 B. 8,961 C. 6,721 D. 4,481.
Hii&ng dan gidi
F T P t f CO the x a y ra:
CI2 + 21- 2cr + I2 (1)
CI2 + 2Br- -> 2C1- + Br2 (2)
N e u p h a n iJng (1) x a y ra h o a n toan, k h o i liTcfng muoi g i a m :
0,2.(127 - 35,5) = 18,3 (g)
K h i ca hai phan ufng (1) va (2) xay ra hoan toan khoi li/png muoi giam:
0,2.(127 - 35,5) + 0,4.(80 - 35,5) = 36,1 (g)
Theo de b a i , k h o i luong muoi g i a m 93,4 - 66,2 = 27,2 (g)
18,3 < 27,2 < 36,1
=> Chufng to phan uCng (1) xay ra hoan toan va c6 mot phan phan ufng (2)
Goi n^^^ pir = X t h i k h o i l i i O n g cua muói giam: 18,3 + x(80 - 35,5) = 27,2
.H^ x = 0,2 (mol) nc,^ = - ( 0 , 2 + 0,2) = 0,2 (mol) =^Ve,^ =4,48(1)
Chon dap an D.
B a i 12: Chat X cd k h o i lupng phan tuT la M . M o t dung dich chat X cd
nong do a mol/1, khói Itrpng rieng d gam/ml. Nong dp C% cua dung dich X la
ạM d.M 10a n
lOd 10a M.d lOOOd
Hii&ng đn gidi
X e t 1 lit dung dich c h a t X :
a-M.lOO ạM
=> nx = a mol->mx = a. M => mad x = = lOOOd => C % = — — .
\j /o lUCl
C h p n dap a n Ạ
B a i 13: D u n g dich chufa a mol N a O H tac dung vdi dung dich chuTa b mol
H3PO4 s i n h r a h o n hop N a 2 H P 0 4 + Na3P04. T i so f l a
b
Ạ 1 < - < 2. B. - > 3. C . 2 < < 3. D . - > 1.
b b b b
HiCdng đn gidi
C a c phiiong t r i n h p h a n iJng:
N a O H + H3PO4 > NaH2P04 + H2O (1)
2 N a O H + H3PO4 > N a 2 H P 0 4 + 2H2O (2)
3 N a O H + H3PO4 — > Na3P04 + 3H2O (3)
T a cd: nNaOH = a mol; n^^pQ^ = b mol.
De thu dupe h o n hpp muoi N a 2 H P 0 4 + Na3P04 t h i p h a n ufng x a y r a d ca h a i phirong t r i n h (2 v a 3), do đ:
2 < i W L < 3, tiJc l a 2 < ^ < 3.
B a i 14: Tron a gam hon hgfp X gom 2 hidrocacbon C 6 H 1 4 va CeHe theo t i
le so mol (1:1) v6i m gam mot hidrocacbon D r o i dot chay hoan toan
, . , » 275a . 94,5a „ m
t h i t h u dirac - — - gam CÔ va gam H 2 O .
82 oZ
a) D thuoc loai hidrocacbon nao
Ạ C„H2n.2. B. CnH2„-2. C. C„H2„. D. C„H„. b) Gia t r i m la
Ạ 2,75 gam. B. 3,75 gam. C. 5 gam. D. 3,5 gam.
HUcfng dan gidi
a) Chon a = 82 gam
Dot X va m gam D (C^Hy) t a c6:
19 275 44 94,5 = 6,25 mol = 5,25 mol C 6 H 1 4 + — O2 > 6 C O 2 + 7 H 2 O CeHe + — O2 > 6 C O 2 + 3 H 2 O Dot D: C H . , 4 -> xCO, + ^ H , 0
Dat n p H = H p H = b mol ta c6: 86b + 78b = 82 => b = 0,5 mol. Dot 82 gam hon hc/p X t h u diroc: n^ô = 0,5.(6 + 6) = 6 mol
n„^o =0.5-(7 + 3) = 5 mol
=> Dot chay m gam D t h u dtfcJc: n c o ^ = 6,25 - 6 = 0,25 mol
n „ p = 5 , 2 5 - 5 = 0,25 mol Do n^Q^ = n„ 0 -> D thupc CnH2n. Chon dap an C.
b) mo = m c + mH = 0,25.(12 + 2) = 3,5 gam.
Chon dap an D.
B a i 15: X la hc(p kirn gom (Fe, C, FeaO, trong do h a m Ixiong tong cong
cua Fe la 96%, h a m liTcfng C dofn chat la 3,1%, h a m lUcfng FesC la a%. Gia t r i a la
Ạ 10,5. B. 13,5. C. 14,5. D. 16.
Hii&ng dan gidi
Xet 100 gam hon hop X ta c6 mc = 3,1 gam, m^^^c = ^ S^^^ va so gam
Fe tong cong la 96 gam.
a = 13,5.
1 9a
• n i c , . „ n . . e 3 C , = 1 0 0 - 9 6 - 3 , l = — Chon dap an B.
Jai 16: H o n hgfp X c6 mot so ankan. Dot chay 0,05 mol h5n hgp X t h u dugc a mol C O 2 va b mol H 2 O . Két luan nao sau day la dung?
Ạ a = b. B. a = b - 0,02. C. a = b - 0,05. D. a = b - 0,07.
Hit&ng dan gidi
Cdch 1: Dat cong thiJc tong quat cua 1 so ankan la C^H^j^^^
3x + l 0,05 O 2 > X C O 2 + (x + D H z O > 0,05x -> 0,05 (x + 1) mol a = b - 0,05. 0,05x = a 0,05 (x + 1) = b Chon dap an C. Cdch 2: nankan = nn.o - " c o , --^ ^ C O j ~ '^HjO ~ I^ankan <::> a = b - 0,05 • Chon dap an C.
B a i 17: Phong dien qua O2 diigc h6n hgp k h i O2, O3 c6 M = 33 gam. Hieu
Lị
suat phan iJng la
Ạ 7,09%. B. 9,09%. C. 11,09%.
Hiidng dan gidi
3 O 2 2 O 3
Chon 1 mol hon hgp O2, O 3 ta c6:
nô = a mol = (1 - a) m o l . 15
D.13,09%.
, 15 1 , =>n^ = 1 - = — m o l =:>n 16 16 1 3 3^ . 32 o, bjoxihod -j^g-g 32 mol .100
Hieu suat phan lirng la: y = 9,09% . 3 2 ^ 16 • Chon dap an B.
B a i 18: (DHKA- 2007): H o n hop gom hidrocacbon X va oxi c6 t i le so
mol ttfOng iJng la 1:10. Dot chay hoan toan h 5 n hop t r e n t h u duge hon hop k h i Ỵ Cho Y qua dung dich H 2 S O 4 dSc, t h u dugc h o n hop k h i Z CO t i khoi doi v d i hidro bSng 19. Cong thufc phan tuf cua X la
Ạ C 3 H 8 . B. C 3 H 6 . C. C 4 H 8 . D. C 3 H 4 .
Hii&ng đn gidi
Dot hon hop gom hidrocacbon X gom C^Hy (1 mol) va O2 (10 mol).
C x H y +
1 m o l
O2 > X C O 2 + - H 2 O
4
y
mol >x mol — mol 2
Hon hop k h i Z gom x mol C O 2 va = 19.2 = 38 1 0 - x + y 4 mol O2 di/. ( " C O , ) 44 (no,) 32 38 Heo, ^ 1 Vay: x = 1 0 - x - ^ - > 8x = 40 - y 4 : ^ X = 4, y = 8 Vay CTPT X la C 4 H 8 . Chon dap an C.
[ B a i 19: A la h o n hop gom mot so hidrocacbon d the k h i , B la k h o n g k h i . Trpn A v d i B d cung nhiet do ap suat theo t i le the tich (1:15) di/gc h6n hop k h i D. Cho D vao binh k i n dung tich khong doi V . N h i e t do va ap suat trong binh la t°^ va p atm. Sau k h i dot chay A trong binh chi c6 N 2 , C O 2 va h o i niTdc v d i V^Q^ : V„^o = 7 : 4 daa b i n h |ve t .
lAp suat trong b i n h sau k h i dot la pi c6 gia t r i la ^- P i = 47 48^ B. p i = p. „ 16 C. P, = — p . 17 D. p . = - p .
Hii&ng dan gidi
)ot A: C x H y + X + -