, rNaHC0j:(2b-a) mol
Phufdng trlnh phSn tfng:
Phan dpng \ii phuong pWp giai Hoa hyc 11 VP CO - DO Xufln Hiftig 2H2 + 0 2 - ^ 2H2O X mol — mol 2 2C0 + O2 — 2 C O 2 y mol ^ mol n„ =-?l^ = 0 , 4 m o l = > - + ^ = 0,4 22,4 2 2 Ta c6 h? phiTcJng trinh : 2x + 28y = 6,8 ^ + y = o. 4 ^ 2 2
* Thanh phan % theo the tich hon hdp :
x = 0,6 y = 0.2 (mol)
%V = M . i 0 0 % = 75%
0,8 %Vco=25%.
* Thanh phan % theo khoi lifdng hon hdp :
%m^^^ = 100% = 17,65% %mco = 82,35%.
Bai 13. Nung n6ng 7,2g Fe203 vdi khi CỌ Sau mpt thdi gian thu drfdc m(g) chS't
rin X. Khi sinh ra hip thu het bdi dung dich BăOH)2 dtf(?c 5,91(g) ket tua, tiep tvc cho Bă0H)2 du vao dung dich tren th^y c6 3,94(g) ket tua nifạ Tinh m.
Giii
HS'p thu C O 2 vao dung dich Bă0H)2 thl c<3 k^t tiia BaCOs xuSft hi^n, cho tigjp dung dich Bă0H)2 du vao lai c6 ket tiia, chtfng t6 C O 2 tic dung vcti
dung djch Bă0H)2 sinh ra '2 mu^ị
CO2 + Bă0H)2 -> BaCOji + H2O (1) 5,91
0,03 = 0,03 mol
(2) 197
2CO2 + Bă0H)2 -> BăHC03)2
0,02 0,01
BăHC03)2 + Bă0H)2 -> 2BaC03 i + 2H2O (3)
• 0,01 TheophSntirngd), (2), (3) TheophSntirngd), (2), (3) 3 94 - - - i ^ = 0,02 mol 197-—^ I n ^ ^ = 0,03 + 0,02 = 0,05 mol CO2
Trong phin iJng khi3r cic oxit b^ng CO, ta lu6n c6:
158
no (trong oxit) = n c o = n^oj =0,05 mol
. m = mp^^oj - mo = 7.2 - 0.05.16 = 6.4 (g)
Bki 14. Cho lu6ng khi CO di qua m (g) Fe203 dun n6ng. thu dir^c 39,2 gam hon
hdp gom 4 chat r^n la s^t kim loai \k ba oxit ciia n6, dong thdi c6 hSn hdp khi thodt rạ Cho hon hdp khi nay hS'p thu vko dung dich niTdc v6i u-ong c6 du
thi thu du^c 55 gam ket tfiạ Gid tri cda m la:
am
55
Tac6: nr •ceo, - = 0,55 mol
CO2 + CăOH)2 -> CaCOji + H2O
0,55 0,55
Trong phdn tfng khijf cic oxit b^ng CO, ta lu6n c6:
no (trong oxit) = n c o = Hcoj =0,55 mol => m = 39,2 + mo = 39,2 + 16.0,55 = 48 (g)
Bki 15. KhiJr hoan toan 8,72 gam hSn hdp X g6m Fe203 va FeO b^ng CO thi thu dtfdc m gam chS't r^n Y va khi CO2. HSp thu hoan toan khi C O 2 b^ng nir<5c
v6i trong dvt thu diiigJc 6 gam k6't tuạ Tinh gid tri ciSa m.
Gidi
HSfp thu C O 2 vao CăOH)2 dvt => chi tao mu^i CaC03.
n c a c o 3= j ^ = 0.06 mol
CO2 + Că0H)2 CaC034' + H2O
0,06 0,06 Ta c6: ncophiniJng = ^coi = 0,06 mol
Ap dung dinh luat bao toan khoi liTdng, ta c6: m^x + mco = my + mcoj
o my = 8,72 + 0,06.28 - 0,06.44 = 7.76 (g)
Bki 16. Cho hoi nude di qua than n6ng d6, thu dugc 15,68 lit h6n hgrp khi X (dktc) gom CO, C O 2 va H2. Cho toan bp X tdc d\ing h6t vđ CiiO (du) nung 'n6ng. thu dugrc h6n hgp chat rdn Ỵ H6a tan toan bp Y bSng dung djch HNO3
(loSng, du) dugrc 8,96 lit NO (san p h k i khu duy nhat, or dktc). Tinh p h ^ trSm ve the tich khi CO trong X.
Gidi
H2O + C — ^ CO + H2
X X X
Phan dgng va pluwng ph&p giai H6a hpc 11 VP cfl - P5 Xuan Hung
2H2O + C CO2 + 2H2