D. (NH2)2CO, NH4HCO3 ,CO va CO
Cflu 8 Sómol NH, =— =0.5 mol
R T PhiWng trinh phan iJng PhiWng trinh phan iJng
4 N H 3 + 5 O 2 > 4 N O + 6 H 2 O 0,5 m o l 0,5 mol 2 N O + O 2 - > 2 N 0 2 0,5 0,5 m o l 4 N O 2 + O 2 + 2 H 2 O - > 4 H N O 3 0,5 0 . 1 2 5 0,5mol mjd = n i H ^ o + + niNô = 173 g C % H N O , ^ • ^ ^ ^ l l O O % = ^ . 1 0 0 % = 1 8 . 2 % m 173 => C h o n B C a u 9. X e l phan dng: 2 N O 2 ± 2 N O + O2 N o n g d o b a n d i l u c u a N O 2 = N o n g d o N O 2 c a n b a n g + N o n g d o N O c a n b a n g = 0,06 + 0,24 = 0,3 mol// => C h o n A
Cfiu 10. G o i X la % c u a N O irong hon hdp. v a y 100 - x lii % ciia N O 2
T a c 6 30x + 46( 100 - x ) = 36.4 x 100
X = 6 0 % => C h o n C => C h o n C
CSu 11. D u n g B a ( 0 H) 2 c o the nhan b i e t diidc c a c ho a chiíl, l i i c n tiTrtng nhiT saii:
- M a u N H 4 C I giiii phong khi miii khai
- M a u M g C b x u a t h i c n k c l I l i a t r t { n g M g ( 0 H) 2
- M a u F e C l j x u a l h i c n k e l l u a nau di) F c C O H ) , M a u F C C I 2 x u a l h i ^ n k c t tua t r a n g xanh F c C O H ) .
M a u A1(N0,)< xuat h i e n k c t l u a iriing A K O H ) . , . n c u c h o du" B a ( O H) 2 k c t l i i a I a n
- M a u C u C l i xuat h i c n k c t tua xanh C u ( 0 H) 2
C h u ỵ- Troni^ tlu/c te mau FeClj co mdu ndu dọ CuCh cd mdu xcmh lam. Do
do ta cd the nhan biet difoc md khdnfi Z'''"' ^"".'>' ^^'"-^ '-'^'^^^ khdc.
=> C h o n C .
cau 12. A p d u n g c o n g thuTc: C M = ^ " ^ ^ " ^ ' " ^ ' ^ = 3 , 4 7 M M
C h o n B
hSn dgng vi phaong ph^p gii\a hpc 11 VP co - D5 Xuan Hung
;:au 13. Cho quy tim h1n lúcJt vao tiTng mau thuT:
Mau lam quy hoa xanh la K2CO3 va NaOH MSu lam quy hoa do la HNO3 va (NH4)2S04 CăN03)2 khong doi mau quy tim
Cho CăN03)2 vao 2 mau lam quy tim hoa xanh, mau tao kct tua la K2CO3 mau con lai la NaOH. Lay NaOH cho vao 2 mau lam quy tim hoa dọ Man
giai phong khi c6 miii khai la (NH4)2S04, con lai la HNO3. Hoac lay K2CO3 cho vao 2 mau Ircn, mau c6 giiii phong khi la HNO3
=> Chon A
;:au 14. Goi cong thtfc ciia muoi nitrat la .M(N03)n 2M(N03)n ' — ^ MzOn + 2nN02 + n/202 a a/2 na na/4 Ta CO ăM + 62n) = 9,4 (1) 0 , 5 a ( 2 M + 1 6 i i ) = 4 (2) ( M + 6 2 n ) 9,4 Lay (1): (2) ta du^dc: M = 32n. M + 8n 4 Khi n = 2 thl M = 64.
Vay muoi nilral la Cu(N03)2 => Chon C
:au 15. Goi X, y la so mol ciia KNO3 va Cu(N03)2 trong hon hdp
PhiTcfng trinh phan uTng xay ra:
2 K N O 3 . — 2x 2x 2Cu(N03)2 2y -> 2KNO2 + O2 2x X - i ^ 2 C u 0 + 4y Ta CO he phiAJng trinh: 4NO2 + O2 y 10L2x + 188.2y = 95,4 32(x + y)4-46.4y x + 5y = 37,82 Giai he tim x, y tiT do chpn két qua dung. => Chon B
'au 16. Phiin ufng nhict phan:
Pb(N03)2 — ^ PbO + 2NO2 + I/2O2
X 2x x/2
Goi X la so mol Pb(N03)2 da nhiet phan:
Khoi iMng khi thoat ra = 2x.46 + 0,5x.32 = 66,2 - 55,4 = 10,8
= > X = 0,1 mol
Vay hieu sual phan iJng la: H = 50% => Chon D o n
:&u 17. The tich khi thoat ra = 2.5x x 22,4 = 5,6 lit Chon A