D. (NH2)2CO, NH4HCO3 ,CO va CO
Al 3e->Ar'
39x 13x
2N^' + 8 e - > 2N*' 16x 2x
Qua trinh nhan e N ^ ' + . 3e ^ N ^ ' 3x x 2N^' + lOê 20x 2x N2 Pa c6: 5x = 2 mol => x = 0,4 mol mAi = 13.0,4.27 = 140,4 g => Chon C.
C3u 21. Phi/dng trinh phan (Jng:
3 M + 4nHN03 -> 3M(N03)„ + nNO + 2nH20 => Chon C
Cau 22. Lap lujin, rut difdc M = 32n M la Cu
=> Chpn C
Phan djing phuBng ph^p gijii H6a hgc 11 VP co - D5 Xuan Himg Cfiu 23. Cu -> Cu(N03)2 ^ Cu(OH)2 CuO
Ta c6: So mol dong ban dau = 3.2/64 = 0,05 mol => ncu = nc„c) = 0,05 mol
= > X = 0,05.80 = 4 gam => Chon A Cfiu 24. Phifdng irinh phan uTng xiiy ra:
3Cu + 8HNO, -)• 3Cu(NO,)2 + 2NO + 4H2O
X 2 x/3
Cu + 4 H N 0 ,- > Cu(N03)2 + 2NO2 + 2H2O
y - 2y
-^..30 + 2ỵ46 Ta c6: x + y = 0,2 mol, mill khac = 38
= > X = 3y => X = 0.15.y = 0,05 mol
=> Vkhi = (2x/3 + 2y).22,4 = 4,48 lit => Chon C Cau 25. PhUitiig liinh phan u^ng:
2x8Al + .30HNỌ, -> 8AI(NO.,).i + 3N2O + I5H2O
3xlOAl + 36HNỌ, ^ 10A1(NỌ,)., + 3N2 + I8H2O
46A1 + 168HNO, -> 46Al(NOj)., + 6N2O + 9N2 + 84H2O
Vijy l i Ic A l : N:0 : NO = 46 : 6 : 9 => Chon B. Cau 26. So mol HNO, phan iJ-ng
= 3 X .so mol A l + so mol NO + 2 x so mol N2O + 2 x so mol N2 = 3. 13. 0,4 + 0,4 + 2.2.0,4 + 2.2.0,4 = 19,2 mol
=> The lich axil la 38,4 l i l => Chon Ạ Cau 27. Tinh difdc so mol = 0.1 mol
so mol NỌr = 0,04 mol va so mol Cu = 0,0375 mol Phifdng irinh phan ỉng dang ion n i l gon:
3Cu + 8 H ^ + 2 N O r- » 3 C u' ^ + 2 N O + 4H20
Tri/ckphanifng 0,0375 0,1 0,04 0,0375 0,1 0,04
X c l l Ic ~ —
3 8 2
Qua l i Ig lhay NO,' diT, Cu va H^phiin iJng hcl so mol NO = 0,025 mol ^ V = 0,56 lil =^ Chon D. Cfiu 28. Ctic phu-cJng Irinh phan I'rng:
3 M + 4nl INO,, -> 3M(N03)„ + nNO + 2nH20
a mol a a n/3
2M + mH2S04 M2(S04),„ + mHj
a mol a/2 am/2
Theo de ra: an/3 = am/2 => n : m = 3 : 2. V4y kirn loiii c6 hoa trj lhay doi
a(M+62.3) 159,21 100 100 , giai ra ta CO M = 56 M la Fe - ( 2 M + 96.2) 2 =^ Chon Ạ
i u 29. Tinh so mol cac chat:
n N a O H= « ' 7 . 0 , 2 = 0,14 ( m o l ) nH, p o 4 =0,1.1 = 0,1 (mol) X ^ t t l l c -11N20H_ = 0:14^1 4
nH3í04 Ól
NaOH + H3PO4 NaH2P04 + H2O 2 N a 0 H + H.,P04 -> Na2HP04 + 2H2O
=>Chpn B.
Cfiu 30. So mol ciic chat:
44
n N a O H = —- l ' l ( m o l )
• tao 2 muói
n H,P04 39,2 = 0,4 (mol)
K 98
Xet t i Ic: - ^ ^ ^ ^ ^ = — = 2,75 => tao 2 mucn
nH.,P04
2NaOH + H3PO4 -> Na2HP04 + 2H2O
2x X X
3NaOH + H.,P04 -> Na,P04 + 3H2O
3y y y
Ta CO 2x + 3y = 1,1 va X + y = 0,4 X = 0,1 va y = 0,3. Tinh khoi Ming cac muói la chon diTctc kcl qua => Chon D
-fiu 31. X c l qua trinh
Ca3(P04)2 1 mol 2551 mol .SiO, ,C,1 ->2P- - ^ P A -).2H3P04 2 mol 5102 mol
Khoi li/c.»ng bol quang la: 2551.310. 100 100 1 90 73 1000
Phan dgng va phuong ph^p giai H6a hpc 11 VP cd - D8 Xuan Hi^ig
Q£Mng3. CACBON - SnJC
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