Xet tile nco 2= nBăOH)2 > 2 do do kettua tan het.

, rNaHC0j:(2b-a) mol

Xet tile nco 2= nBăOH)2 > 2 do do kettua tan het.

=> Chon D.

Cfiu 6. Khoi liTdng C nguyen chat = 10.96% = 9,6g, so mol C = 9,6/12 = 0,8 mol

PhiTdng trinh phan drng C + O2 ^ C O 2

So mol C O 2 = 0,8 mol, so mol Bă0H)2 = 0,5mol.

Xdt t i 1$ nC02: nBă0H)2 = 1.6 nen xky ra 2 phan tfng sau day

CO2 + BăOH)2 -> BaCOj + H2O

X X X

2CO2 + Bă0H)2 BăHC03)2

2y y y

Ta CO h^ phOtfng trinh: x + 2y = 0,8 x + y = 0,5

Gidi ra ta dtfcfc x = 0,2 y = 0,3

Khoi liTdng BaCOj = 0,2.197 = 39,4g. => Chon Ạ

Cfiu 7. S6' mol ket tua = 0,0Imol, so mol Bă0H)2 = 0,02 mol

Tri^ng hffpl. So mol C O 2 chi du ta 3 0,01 mol ket tua

CO2 + CăOH)2 -> CaCOj + H2O

0,01 mol 0,01 mol The tich C O 2 = 0,01.22.4 = 0,224 lit

rrU&ng h^p 2. So mol C O 2 dir h6a tan m6t phan ket tiia

CO2 + BăOH)2 -> BaCOj + H2O

0,01 0,01 0,01

2CO2 + Bă0H)2 -> BăHC03)2

2y y y 162

Ta C O 0,01 + y = 0,02 =>y = 0,01

Tong so mol C O 2 = 0,01 + 0,02 = 0,03 mol The tich C O 2 = 0,03.22,4 = 6,72 lit

Vay chon C. Cfiu 8: Cdc phan urng

Fe304 + 4 C 0 — 3 F e + 4 C O 2

CuO + CO — ^ Cu + C O 2 C O 2 + Că0H)2 > CaCO, + H 2 O

Ho (trong oxio = "co = n

f^Q^

0,05

m„xii = niKi + m, „ i , r „ n g o x i t = 2,32 + 0,05.16 = 3,12 (g). => Dap an Ạ Cfiu 9: The tich oxi trong khong khi da dung = 20%. 11,2 = 2,24 lit

PhU'dng trinh chay 2 C O + O2 ^ 2CO2

The tich CO = 2,24.2 = 4,48 lit => %C0 = (4,48/8,96). 100%. = 50% =:>DdpanB.

Cfiu 10: Khi cho dung dich HCl tiif tir vao dung djch NazCOj, xdy ra phdn xtng

theo trinh tiT sau:

HCl + Na2C03-> NaHCOj + NaCl ( I )

b b b

HCl + NaHCOj - > NaCl + CO2 + H2O (2) (a - b) (a - b)

Sau phan ifng, cho dung djch Că0H)2 diT vSo dung dich X c6 ket tiia, chiirng to sau phan tfng (2) NaHCOj diT => HCl het => tinh the tich C O 2 theo HCl.

Theo phan tog (2) => n^ô = (a - b) mol V ^ Q J = 22.4(a - b) (Ut) => Dap dn Ạ 2Fe203 + 8SO2 (1) 0,3 Cfiu 1 1 : Ptptf: 4 F e S 2 + I I O 2 0,15 T a c 6 nB„oH)2 = 0,15 mol nKOH = 0 , l mol n 2^ = 0,15 mol n ^ ^ . = 0 , 4 m o l

Khi cho SO2 vao dung djch X thu diTdc 21,7 (g) - » BaS03. Cho Y i&c dung

vdi dung djch NaOH thay xuát hien them ket tiia, chtfng to trong dung dich Y c6 ion HS03'.

han dgng phuong ph^p giSi H6a hgc 11 VP CO - D8 Xufln Hung

T a c 6: n B a c o , = 0,1 mol

Ptptf: SO2 + 2 0 H - > SOi^- + H2O (2) 0,1 0,2 0,1

SO2 + OH" > HSO3" (3)

0.2 0.2

Bâ* + SOj^- > BaS03 i 0,1 0,1

Tac6: n _ = 0,4 - 0.2 = 0,2 mol

OH

Theo ptptf (2), (3) ta c6: n^ô = 0.1 + 0,2 = 0.3 mol Theo ptptf (1) ta c6: rip^s^ = "J^soj = 0.15 mol

=> mpcsj = 120.0,15 = 18(g) =>Ddp an C.

BAI T A P Ti; G l A l

)ai 1. Viét cic phifdng trlnh hoa hoc cua phan iJng bieu dien sd do chuyen h6a sau :

a) C > CO2 > CO > FejOi > Fe(N03)3 > FezOj b) CO2 > NuiCOi > Na2Si03 > HaSiOj > SiOz

c) CO2 > CaC03 > CăHC03)2 > CO2 > C > C > CO2

d) Si > SiOz > K2Si03 > K2CO3 > K C l

/ \

SiF4 CO ).C0Cl2.

)ai2.

a) U m the n^o de phan biet khi CO2 khi O2

b) L ^ m thd' n^o d^ phan bipt muoi natri cacbonat va muoi natri sunfit?

Hudng dJn :

a) Dilng que dom c6n than hong bung chay -> O2.

b) Cho tac dung vdi axit HCl sau do dan san pham qua nÚdc brom -> Na2S03.

[Jai 3. Cho c&c chát: silic, natri silicat, natri cacbonat, magie silixua, axit silixic. Hay lap th^nh mpt day chuyen hoa giffa cac chát tren va viet cac phiTdng trinh hoa hoc.

Sal 4. Khi nung mot hon hdp gom cat tr^ng va than coc trong 16 di^n den 3500"C. thi thu di/dc mot hdp chat chufa khoang 70% Si va khoang 30%C. Viet phiTdng trlnh h6a hoc cua phan ỉng do, biet r i n g mot trong c^c s^n pham cua phan ỉng la cacbon mono oxit.

Bip so : Hdp chat Ih SiC.

Ski 5. Xdc dinh the tich hidro (dktc) thoit ra khi cho Irfdng dtf dung dich natri hidroxit t^c dung vdi mot hon hdp thu duTdc b i n g cAch nau chay 6g magie vdi 4,5g silic dioxit. Gia sijT phan tfng di/dc tien h^nh vdi hi^u suát 100%.

Bips6: VHJ =3,36 lit.

I HUcJfnfi dan: 2Mg + Si02 2MgO + Si

Si + 2NaOH + H2O > Na2Si03 + 2H2t.

Bai 6. Cho dung dich mu6'i CăHC03)2 tdc dung vdi cdc chS't sau : K2CO3, BăN03)2, NaOH, HNO3, Că0H)2, BăOH)2. Viet phi/dng trlnh phan urng xay ra (néu c6).

Biki 7. B i n g phiTdng phap hoa hoc hay nhan b i e t :

a) CaCOa, Na2C03, Na2S04, CaS04.2H20 (dilng ntfdc v^ dung dich HCl). b) Dung dich : NaOH, Na2S03, CaCb, CăHC03)2, Na2S04.

Bai 8. Hoan th^nh cac phtfdng trlnh hoa hoc sau :

1. M g + Si > 2.CaO + C - ^ ^ ? + C 0 3. Fe304 + CO ? +? 4. Si + F2 >?

5. Si + K O H + ? >? + H2T 6. CO2 diT + Bă0H)2 > 7. C + KCIO3 > 8. C + S >

9. C + H2S04dH- )-?+? + H20 10. Na2SiO3 +? + H2O >? + NazCOj.

Bai 9. Khi nung 97,7 gam hon hdp X gom NH4HCO3, NaHC03 v^ CăHC03)2 den khoi li/dng khong doi thu difdc 32,4 gam chat r^n Ỵ Cho chát r^n Y tic

I dung het vdi dung dich HCl lay diT, thu diTdc 4,48 lit khi (dktc). X i c dinh • thanh phan % cac muoi trong hon hdp.

Dap so: %NH4HC03 = 32,37%; %NaHC03 = 34.43%; %CăHC03)2 = 33.2%.

Hudn}; dan : NH4HCO3 -A NH3 + CO2 + H2O 2NaHC03 NazCOj + CO2 + H2O CăHC03)2 CaO + 2CO2 + H2O R^n Y gom Na2C03 \h CaỌ

hki 10. Khur a gam mot oxit s^t bkng khi CO d nhi$t dp cao thu diTdc 0.84g s^t

\h 0.88g CO2.

a) Xac dinh cong thtfc oxit s^t da diJng.

b) Tinh the tich dung dich HCl 2M can de h6a tan het a gam oxit s^t ndi tren. Daps6': a) Fe304 b)VHci = 20ml.

Phan dgnp vi phuong ph^p giai H6a hpc 11 VP co - P8 Xuan Hung

Bai 11. Cho 2,464 lit khi C O 2 (dktc) vao dung dich NaOH, sinh ra 1 l,44g hon hcJp hai muóị Xac dinh khoi ItfcJng moi muói trong hon hdp thu dtfdc.

hcJp hai muóị Xac dinh khoi ItfcJng moi muói trong hon hdp thu dtfdc.

Ddp só:

= lO-^g ; mNaHco, = 0.84g.