- VI, Fe co nd => AgN03,Cu(N03)2 phai phan ling hét Dodo dung dic hX chua A1(N03)3 va Fe(N03)
FeO B CrO D Cr203 D Fe
Phdn loai va phuan^ phdp gidi nhanh ET H6a hoc 12 -Cu Thanh Toan Hu&ng dan gidi
'j Theobaira: 11^0 = 0'8mol;nso2 = 0, 9 m o l . ,.j,c;,i! K< iwi O M U
- V\o nso2 ^ ' " ^ ^ ^^^^ (trong oxit c6 s6' oxi hod thap, trong mu6'i sunfat C O so oxit hod cao). ; " ' - Neu trong oxit, s6' oxi hod ciia M la +2, trong mu6'i sunfat I6n +3 thl khi nco = 0,8 " moltm n^ 2 ^ | . n c o = U m o l . , .^j^) ^pfcH£ 4 ^ 1 ^ , '
Nhung theo bai ra nso2 = 0'9 < 1.2
=> Chi m6t phdn kirn loai M thay d6i s6' oxi hod. Vay oxit phu hop la Fe304. Dap an dung la D.
Dang 13: Bai tap ve dong va hop chat ciia dong 1. Truong hop xay ra phan ufng la
Ạ Cu + Pb(N0,)2 (loang) B. Cu + HCl (loang) C. Cu + HCl (loang) + O, ^ D. Cu + H2SO4 (loang)
Hu&ng đn gidi
Hoan thanh cdc PTPU" (neu c6):
Cu + PKNOi), >kh6ngxayrạ Cu + HCl > khong xay rạ
2Cu + 4HC1 + O, > 2CuCl, + 2 H , 0 Cu + H 3 S O 4 (1) > khong xay rạ
Dap an dung la C.
2. Cho 0,448 lit khi N H , (dktc) di qua 6'ng su: dirng 16 gam CuO nung nong, thu duoc chat ran X (gia sir phan ufng xay ra hoan toan). PhSn trdm kh6'i luong ciia duoc chat ran X (gia sir phan ufng xay ra hoan toan). PhSn trdm kh6'i luong ciia Cu trong X la
Ạ 12,37% B. 14,12% C. 85,88% D. 87,63%
,( Huong đn gidi
Theobai la: nf^^^ =0,02mol; UQ^^Q = 0,2mol PTPIT: 2NH3 + 3 C U O >3Cu + N 2 + 3 H 2 0 Banđu: 0,02 0,2 Phan ling: 0,02 0,03 -> 0,03 , Con: 0 0,17 0,03 0,03.64.100% vay %mp,,,x = = 12,37% 0,03.64 + 0,17.80 Dap an dung la Ạ
Cty TNHH MTV DVVH Khang Viet
3. Chát ran X phan iJng \6\g djch HCl dugc dung dich Ỵ Cho tir t\x dung djch NH, den du vao dung djch Y, ban đu xuát hien kfit tua xanh, sau do kS't tiia tan, NH, den du vao dung djch Y, ban đu xuát hien kfit tua xanh, sau do kS't tiia tan, thu duoc dung djch mau xanh thdm. Chat X la
Ạ Cu B. CuO C. Fe ' D. FeO Hu&ng đn gidi Chát X la CuO ^CuCl2 + H20 CuO + 2HCl ( X ) ( d d Y )
2NH3 + CUCI2 + 2H2O ^ C u ( O H) 2 i +2NH4CI (đY) xanh ^ C u ( N H, ) J( O H) 2 >1: S i . ' 4 N H 3 (d) +Cu(OH) f xanh thSm ? Dap an diing la B.
I 4 . Cho cac dung djch loang: (1) FeCl,, (2) FeCU, (3) H,S04, (4) HNO,„ (5) h6n hop gom HCl va NaNOj. Nhung dung djch phan Aig duoc vdi kim loai Cu la
A . ( l ) , ( 2) , ( 3 ) B . ( l) , ( 3) , ( 5 ) C . ( l) , ( 3) , ( 4 ) D. (1), (4), (5)
Hu&ng đn gidi
PTHH: ' ' ' (I) 2FeCl3 + Cu^2FeCl2 + CUCI2
(4) 8 H N O 3 + 3Cu -> 3Cu (NO3 \ 2N0 + 4 H 2 O
(5) 8HCl + 2 N a N 0 3+ 3 C u ^ 3 C u C l 2+ 2 N a C l + 2NO + 4H20 ^1
Dap an diing la D. 5. Nung 6,58 gam Cu(NO,): trong binh kin khong chua khong khi, sau m6t thcri
gian thu duoc 4,96 gam chat rdn va h6n hop khi X. Hap thu hoan toan X vao n-
U6c de duoc 300 ml dung djch Ỵ Dung djch Y c6 pH bang
Ạ 2. B. 3. C.4. D. 1. ; <
Hu&ng đn gidi
Cu(N03), (riri) CuO(,i„) + 2NO,t+ - 0, t 2
x(mol) 2x 0,5x Khoi luong chát rdn giam bang khoi luong cua h6n hop khi NO, va O,:
(hh kht) = 6,58 - 4,96 = 1,62 = 2x . 46 + 0,5x . 32 x = 0,015 (mol). 1 2 N 0 , + - 0 , + H , 0 ' 2 ' 2x 0,5x -> 2HNO3 2x (mol)
n„N0 3 = 2x = 2. 0,015 = 0,03 (mol). [H^] = [HNO,] =0,1 (M).
pH = - l g [ H l = l . , ,
Dap an dung la D. 6. Hoa tan m gam Cu vao 0,2 lit dung dich X g6m HNO3 0,2M va H.SÔ 0,2M,
phan ling tao ra san phdm khu duy nhát la NỌ Gia tn lorn nhát cua m la Ạ 0,96 gam. B. 4,8 gam. C. 2,88 gam. D. 6,4 gam.
Hu&ng dan gidi ,
Theo bai ra: So mol HNO3 = 0,2 . 0,2 = 0,04 (mol)
So mol H,S04 = 0,2 . 0,2 = 0,04 (mol). " => n + = 0,04 + 0,04 . 2 = 0,12 (mol); n _ = 0,04 (mol). PTPlT: 3Cu + 2NO3 + 8H^ — 0,36/8 0,12 (mol) 0,12.3 3Cu-" + 2 N 0 t + 4H2O 8 0,12.3.64 , mc„ = 2,88 (gam) 8 ... 1; Dap an dung la C. 7. Thirc hien hai thi nghiem:
1. Cho 3,84 gam Cu phan irng vdri 80 ml dung dich HNO3 I M thoat ra V, lit NỌ 2. Cho 3,84 gam Cu phan ufng v6i 80 ml dung dich chiia HNO3 I M va H2SO4 0,5M
thoat ra lit khi NỌ
Biet NO la san phdm khu duy nha't, cac thd' tich khi do a cung di6u kien.
Quan ht giiJa V, va V, la
A . V , = 1,5V,. B.V2 = 2V,. C.V, = 2,5V,. D.V2 = V,.
Hu&ng đn gidi
Theo bai ra: ncu = 0,06 mol
* Thi nghidm 1: n^,No, = 0,080. 1 =0,08 (mol). PTPlT: 3Cu + 8HNO3 —
Ban dSu 0,06 0,08 (mol) Phan umg: 0,03 <- 0,08
0,08.2
3Cu(N03)2 + 2NO +4H2O 0,02 (mol) Suy ra: nNO(i) =
8 = 0,02 (mol) * Thi nghiem 2: O H N O J = 0,08 mol;
nH 2 S O 4 = 0.08. 0,5 = 0,04 (mol) nNô = 0,08 mol; n ^ = 0,08 + 0,04. 2 = 0,16 (mol). H fMiiVU^r -y-i , a r ; M l
FTPLT: 3Cu + 2 N O 3 + 8 H * 3Cu-" + 2NO + 4H2O
|p Banđu 0,06 0,08 0,16 ,
Phanung: 0 , 0 6 0 , 0 4 -> 0,16 0,04 (mol) Suy ra: n N O , : , = 0,04 (mol).
Dodo: nNO(2) = 2 n N O( i )
=> V, = 2V| (D6'i vdri chat khi, trong cung difiu kidn, ti 16 v^ s6' mol cung bang ti
\t ve \hi tich). .<y
M Dap an dung la B .