- ,rj Na2C03—^ Kh6ng c6 phan ling (1)
2NaHC03 — ^ NâCOj + CÔ t + H,0 (2)
Theo (2): S6' mol NaHCO, = 2 s6' mol CÔ = 2 . — = 0,2 (mol).
22,4
mNaHco,= 0.2.84 = 16,8 (gam).
%ni,,„eo3 = ^ . 1 0 0 % = 84%; %m^,^^^^ =16%.
2. Dap an diing la B.
Theo (2): S6 mol NajCOj = s6' mol CO2 = 0,1 (mol). ^ niN,,co, = 0,1.106 = 10,6 (gam). Khfíi lugng chát ran X la:
mx = 10,6 + (20 -16,8) = 13,8 (gam).
3. Dap an diing la D.
S6' mol NâCOj = — = 0,13 mol < s6' mol CăOH)^ = 0,2 (mol)
106
Co 1 mu6'i duy nhát la CaCO, duoc tao th^nh:
CO, + Ca (OH)^ du ^ CaCO, i + H^O (3)
Theo (3): S6' mol CaCO, i = s6 mol CO2 = 0,13 (mol) ^
->nic3co,= 0.13.100 = 13 (gam).
2. TL; LUAN
a) Cac phuong trinh hoa hoc:
2Fe + 3Cl2—i^2FeCl3 ? (1)
Cu + Clj—^CuCl^ (2) . FeCl,+3NaOH->Fe(OH),i + 3NaCl (3) . FeCl,+3NaOH->Fe(OH),i + 3NaCl (3)
CuCl^ + 2NaOH -> Cu (OH)^ i + 2NaCl (4)
b) Goi X va y la s6' mol Fe va Cu trong h6n hop.
Theo (1) va (2): S6' mol FeCl, = x, so mol CuClj = ỵ i
Dodo: 162,5.x + 135.y= 18,35 (I)
Theo (3) va (4): S6' mol Fe(OH)3 = x, s6' mol Cu(0H)2 = ỵ
Do do: 107x + 98y = 12,90
Hay: l,09x + y = 0,13 (II) Giai he phuomg trinh (I) va (II) dugc: x = 0,05; y = 0,08. Giai he phuomg trinh (I) va (II) dugc: x = 0,05; y = 0,08.
mp^ =0,05.56 = 2,80 (gam)" m^, =0,08.64 = 5,12 (gam)" m = 7,92 (gam) m^, =0,08.64 = 5,12 (gam)" m = 7,92 (gam) 2 80 %mp = - ^ . 1 0 0 % = 35%. ^ ?s. • 7,92 v . V %mr, = 100% - 35% = 65%. De so 2.4 1. TRAC NGHIEM KHACH QUAN
C a u 1. Dap an diing la B. Cac phuong trinh hoa hoc:
X,C03+2HC1^2XC1 + C0, t + H^O ' (1)
Y C 0 3+2HC1^YC1,+ C O 2 t + H^O (2)
T h e o ( l ) v a(2): n „ , =n„,o = ^ = 0,2(mol) ^ •
22,4 I
Theo dinh luat bao loan kh6'i lucfng: 20 + 36,5.0,4 = m + 44.0,2 +18.0,2 ^ m = 22,2 (gam).
Cau 2. Dap an diing la B.
Goi kim loai hoa tri I I I la M . Phucfng trlnh hoa hoc:
2M + 3 C l 2^ 2 M C l 3 2Mgam 2(M + 3.35,5) gam ' ' ' ' 10,8 gam 53,4 gam 2 M 2 ( M +3.35.5) Ta C O ti 16: = — • 10,8 53,4 106,8M = 21,6(M +106,5) = 27 (Al).
Vay kim loai da diing la A l .
Cau 3.
1. Dap an diing la C.
Tinh so mol cac chát va khÓi luong cac nguyfin tÓ trong X :
n ^ = ^ 0,09 (mol) -> mc = 0,09.12 = 1,08 (gam) - 22,4 n 81 n„,„= — = 0,045 ( m o l ) ^ m „ =0,045.2.1 = 0,09 (gam) • 18 ^ m c + m „ =1,08 + 0,09 = 1,17 (gam) = mx.
' Nhu vay trong thanh phin phan \\i cua hop chát huxi co X g6m cacbon vahidrọ
2. C6ng thirc dung la B.
Dat CTPT ciia X la Q H y (x, y la nhiing s6' nguyen dirong).
Ta c6: X : y = nc : H H = 0,09 : 0,09 = 1 : 1. -> COng thiic don gian nhát ciia X la CH. CTPTla (CH)_^= 13n. 78 13 vay CTPT cua x la Q^^. 2. T i ; LUAN 8 96
SÓ mol hOn hcfp: n^^ = — — = 0,4 (mol). 22,4
a) Cho hOn hop khi vao binh dung dung dich nu6c brom, xay ra phan ung:
C H 2 = C H 2+ B i i^ C H 2 B r - C H 2 B r (1)
CH = CH + 2 B ^ ^ C H B i i - C H B ^ (2) =78 gam = 1 3 n ^ n = — = 6.
b) Goi X va y la s6' mol C2H4 va C2H2 trong h6n hop.
KhÓi luong binh dung djch brom tang 11 gam, chinh la khÓi lugng cua 8,96 IfthOnhgfp. Theo đ bai ta c6: ' x + y = 0,4 28x + 26y = l l 0,3 x = 0,3; y = 0,l % Vc^„^ = . 100% = 75%; %Vc^„, = 25%. u, 4 De so 2.5
1. TRAC NGHIEM KHACH QUAN
Cau 1. Dap an dung la B.
8 96
só mol hOn hop = — — = 0,4 (mol). 22,4
So mol etilen = 0,3 (mol); SÓ mol axetilen = 0,1 (mol). Phuong trlnh hoa hoc:
C^H, + 3O2 0,3 mol 3.0,3 C2H2 + 2,50^ -> 2CO2 + 2 H 2 O 2.0,3 mol ^ 2CO2 + H j O
0,1 mol 0,25 mol 0,2 mol ^ n „ , =3.0,3 + 0,25 = l,15(mol)
X,^ =1,15.22,4 = 26,76 (lit). ^ n^o, = 2.0,3 + 0,2 = 0,8 (mol)
m^o, =0,8.44 = 35,2 (gam).
Cau 2. Dap an diing la Ạ
Dat CTPT cua hidrocacbon X la C,Hỵ
Phuong trlnh hoa hoc:
Theo PT: Theo de bai: 1 lit 1 lit y 4 ' 2 ' xlit 2 lit ^ l i t 2 2 lit - ^ | = l - > x = 2; ^ = l ^ y = 4. vay CTPT cua X la C2H4. (1) (2) • J u
Cau 3. Dap an dung la C.
Thi tich cua 12 kg rugu etylic nguydn cha't la: '
V , = —- 1 5 ( l i t ) .
0,8
Rucm 30" - nghia la trong 1 lit rucru loai nay c6 0,3 lit rupu etyl nguySn chat. Do do, thd tich rugu 30" la:
V „ = 1 ^ = 50 (lit). Ririni(30°) 30
cau 4. C6ng thiic dung la C. ^|
Dat CTPT ciia X la C.H^. ^^
Phirong trinh phan ung chay:
4 O . - ^ x C O j + ^ H ^ O (1)
Theo de bai va theo (1): n^,, = „ = = 2
X 1
^ - = - -> Cong thiic don gian nhát cua X la CHj.
CTPT cua X la ( C H , ^ n = 2-> X la C^H,.
Cau 5. Dap an dung la D.
CH,COOH + C^HjOH < > CH.COOC^H, + H^O
(A) (B)
cau 6. Dap an dung la Ạ
Cac phuong trinh phan ling d^y du: C,H, + CI2 — ^ ^ C , H , C 1 + HCl C,H, + Br, ') C,H,Br,
2. T i ; LUAN
Phuong trinh hoa hoc:
CHjCOOH + C^H^OH < = ^ ^ ^ ^ CHjCOOCjH^+Hp S6' mol cac chat:
ncH,cooH = ^ = 0'02(mol); n^^^,,,, = ^ = 0,025(mol).
n^HjOH > ncH,co()H C^H^OH dụ S6 mol CHjCOOC^H, = 0,02 (mol).
Khd'i lugng este CHjCOOC^H, thu dugc vdfi hieu suat 55% la: m.,„ = 0,02.88.— = 0,968 (gam).
100
Phan IV
GIOI THIEU DE THI HOC SINH G161 LOP 9 - MOIM HOA HOC LOP 9 - MOIM HOA HOC
..''rti.j De so 1
(Thcfi gian lam bai 150 phut)
Cau 1 ( 4 diem)
1. Co 4 dung dich khong mau bi mat nhan: K^SÔ; K^CO,; HCl; BaCl^. a) Chi diing th6m mdt kim loaị a) Chi diing th6m mdt kim loaị
b) Khdng dung them thu6'c thCr nao khac.
Neu each nhan ra tOmg dung djch, viS't cac phuong trinh hoa hoc.
2. Viet cac c6ng thiic cáu tao ling vdi c6ng thiic phan tur C^H^Cl^ ma 6 m6i nguyen tii cacbon khong chiia qua 1 nguySn tur CỊ m6i nguyen tii cacbon khong chiia qua 1 nguySn tur CỊ
3. Hoa tan m6t mu6'i cacbonat ciia kim loai M bang m6t luong vira dii dung dich H2SO4 9,8% thu dugc dung dich mu6'i sunfat 14,18%. Tim kim loai M . dich H2SO4 9,8% thu dugc dung dich mu6'i sunfat 14,18%. Tim kim loai M .
Cau 2. (2 diem)
Láy thi du bang phuong trinh phan ling cho cac trucmg hgp sau:
1. Khi cho mdt kim loai vao mot dung dich mu6'i, san ph^m tao thanh la: a) muoi + kim loaị a) muoi + kim loaị
b) mudi + bazg + khị . ., / • • " -
c) hai mudị • d) duy nhát mdt mudị
2. Khi cho mdt oxit vao a) nude-> axit + oxit a) nude-> axit + oxit b) axit-> hai m u d i + . . . c) kiem ^ hai m u d i + . . . Cau 3. (4 diem) J Cho so do sau: 1 X - > A ^ B ^ C - > D ^ E QH.aO, F G Biet: X la m6t chát khị
A la m6t polime c6 kh6'i luomg phan tir Idn.