\(C+O_2\xrightarrow[]{t^o}CO_2\\ C+CO_2\xrightarrow[]{t^o}2CO\\ 2CO+O_2\xrightarrow[]{t^o}2CO_2\\ CO_2+NaOH\rightarrow NaHCO_3\\ NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)

a)
$CO + FeO \xrightarrow{t^o} Fe + CO_2$
$Fe_2O_3 + 3CO \rightarrow 2Fe + 3CO_2$
$PbO + CO \xrightarrow{t^o} Pb + CO_2$
$CuO + CO \xrightarrow Cu + CO_2$
b)

`CO_2 + 2NaOH -> Na_2CO_3 + H_2O`

`CO_2 + 2KOH -> K_2CO_3 + H_2O`

`CO_2 + Ca(OH)_2 -> CaCO_3 + H_2O`

`CO_2 + Ba(OH)_2 -> BaCO_3 + H_2O`

CuO + CO --t^o--> Cu + CO₂

Ta có: \(n_{O\left(oxit\right)\left(pư\right)}=\dfrac{m_O}{16}=\dfrac{m_{giảm}}{16}=\dfrac{18,56-15,232}{16}=0,208\left(mol\right)\)

Mà \(n_{O\left(bđ\right)}=4n_{Fe_3O_4}=4.\dfrac{18,56}{232}=0,32\left(mol\right)\)

=> \(H=\dfrac{0,208}{0,32}.100\%=65\%\)

=> Chọn A

17. Gọi \(\left\{{}\begin{matrix}n_{NaHCO_3}=a\left(mol\right)\\n_{Na_2CO_3}=b\left(mol\right)\end{matrix}\right.\)

=> 84a + 106b = 27,4 (1)

BTNT Na: \(n_{Na_2CO_3\left(sau\right)}=\dfrac{1}{2}n_{NaHCO_3}+n_{Na_2CO_3\left(bđ\right)}\)

=> \(0,5a+b=\dfrac{21,2}{106}=0,2\left(2\right)\)

Từ (1), (2) => a = 0,2; b = 0,1

=> \(m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

=> A

19. \(\left\{{}\begin{matrix}n_{C\left(CO_2\right)}=n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\n_{C\left(CaCO_3\right)}=n_{CaCO_3}=\dfrac{10}{100}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(n_{C\left(CO_2\right)}>n_{C\left(CaCO_3\right)}\)

=> Kết tủa tan ra một phần

BTNT C: \(n_{Ca\left(HCO_3\right)_2}=\dfrac{n_{CO_2}-n_{CaCO_3}}{2}=0,1\left(mol\right)\)

BTNT Ca: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}+n_{Ca\left(HCO_3\right)_2}=0,2\left(mol\right)\)

=> \(V=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

=> Chọn B

14. Ta có: \(n_{Fe}=\dfrac{13,44}{56}=0,24\left(mol\right)\)

PTHH: \(Fe_3O_4+4CO\xrightarrow[]{t^o}3Fe+4CO_2\)

                           0,32<---0,24

=> \(V_{CO}=0,32.22,4=7,168\left(l\right)\)

=> Chọn C

15. \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\n_{NaOH}=\dfrac{164.1,22.20\%}{40}=1\left(mol\right)\end{matrix}\right.\)

Xét \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1}{0,25}=4>2\) => Tạo muối Na₂CO₃, NaOH dư

BTNT C: \(n_{Na_2CO_3}=n_{CO_2}=0,25\left(mol\right)\)

=> m_muối = 0,25.106 = 26,5 (g)

=> Chọn A

16. Vì nung thu được thêm kết tủa => Kết tủa tan ra một phần

Ta có: \(\left\{{}\begin{matrix}n_{CaCO_3\left(thêm\right)}=\dfrac{0,5}{100}=0,005\left(mol\right)\\n_{CaCO_3}=\dfrac{2}{100}=0,02\left(mol\right)\end{matrix}\right.\)

BTNT Ca: \(n_{Ca\left(HCO_3\right)_2}=n_{CaCO_3\left(thêm\right)}=0,005\left(mol\right)\)

BTNT C: \(n_{CO_2}=2n_{Ca\left(HCO_3\right)_2}+n_{CaCO_3}=0,03\left(mol\right)\)

=> V = 0,03.22,4 = 0,672 (l)

=> C

11. Ta có: \(n_{CO_2}=\dfrac{2,464}{22,4}=0,11\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{Na_2CO_3}=a\left(mol\right)\\n_{NaHCO_3}=b\left(mol\right)\end{matrix}\right.\) => 106a + 84b = 11,44 (1)

BTNT C: \(n_{CO_2}=n_{Na_2CO_3}+n_{NaHCO_3}\) => a + b = 0,11 (2)

Từ (1), (2) => a = 0,1; b = 0,01

=> \(m_{NaHCO_3}=0,01.84=0,84\left(g\right)\)

=> Chọn B

12. BTNT C: \(n_{CO_2}=n_{NaHCO_3}=\dfrac{37,8}{84}=0,45\left(mol\right)\)

=> \(V_{CO_2}=0,45.22,4=10,08\left(l\right)\)

=> Chọn D

13. Ta có: \(n_{Fe}=\dfrac{3,36}{56}=0,06\left(mol\right)\)

=> \(n_{Fe_xO_y}=\dfrac{1}{x}.n_{Fe}=\dfrac{0,06}{x}\left(mol\right)\)

=> \(M_{Fe_xO_y}=\dfrac{4,8}{\dfrac{0,06}{x}}=80x\left(g/mol\right)\)

=> 56x + 16y = 80x

=> \(\dfrac{x}{y}=\dfrac{16}{80-56}=\dfrac{2}{3}\)

=> CTHH của oxit là Fe₂O₃

Vậy 13 chọn B

9. Ta có: \(\left\{{}\begin{matrix}n_{CO_2}=\dfrac{10,08}{22,4}=0,45\left(mol\right)\\n_{CaCO_3}=\dfrac{35}{100}=0,35\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}n_{C\left(CO_2\right)}=n_{CO_2}=0,45\left(mol\right)\\n_{C\left(CaCO_3\right)}=n_{CaCO_3}=0,35\left(mol\right)\end{matrix}\right.\)

=> \(n_{C\left(CO_2\right)}< n_{C\left(CaCO_3\right)}\)

=> Kết tủa bị hòa tan một phần

BTNT C: \(n_{Ca\left(HCO_3\right)_2}=\dfrac{n_{CO_2}-n_{CaCO_3}}{2}=0,05\left(mol\right)\)

BTNT Ca: \(n_{Ca\left(OH\right)_2}=n_{CaCO_3}+n_{Ca\left(HCO_3\right)_2}=0,4\left(mol\right)\)

=> \(V=\dfrac{0,4}{8}=0,05\left(l\right)=50\left(ml\right)\)

=> Chọn B

10. BTNT Ba: \(n_{BaCl_2}=n_{BaCO_3}=\dfrac{108,35}{197}=0,55\left(mol\right)\)

ĐLBTKL:

\(m_{Na_2CO_3,\left(NH_4\right)_2CO_3}+m_{BaCl_2}=m_{BaCO_3}+m_{NaCl,NH_4Cl}\)

=> \(m=m_{NaCl,NH_4Cl}=55,8+0,55.208-108,35=61,85\left(g\right)\)

=> Chọn C

Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Cu}=b\left(mol\right)\end{matrix}\right.\) => 56a + 64b = 8,64 (*)

\(n_{khí.B}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

Ta có: \(\overline{M_B}=19.2=38\left(g/mol\right)\)

=> \(m_B=0,16.38=6,08\left(g\right)\)

Gọi \(\left\{{}\begin{matrix}n_{NO}=x\left(mol\right)\\n_{NO_2}=y\left(mol\right)\end{matrix}\right.\) =>  \(\left\{{}\begin{matrix}x+y=0,16\\30x+46y=6,08\end{matrix}\right.\)

=> x = y = 0,08 (mol)

Quá trình oxi hóa:

\(Fe^0\rightarrow Fe^{+3}+3e\)

a--------------->3a

\(Cu^0\rightarrow Cu^{2+}+2e\)

b--------------->2b

=> \(\sum n_{e\left(nhường\right)}=3a+2b\)

Quá trình khử:

\(N^{+5}+1e\rightarrow N^{+4}\)

0,08->0,08

\(N^{+5}+3e\rightarrow N^{+2}\)

0,08->0,24

=> \(\sum n_{e\left(nhận\right)}=0,08+0,24=0,32\left(mol\right)\)

Bảo toàn electron: \(\sum n_{e\left(nhận\right)}=\sum n_{e\left(nhường\right)}\)

=> 3a + 2b = 0,32 (**)

Từ (*), (**) => a = 0,04; b = 0,1

=> \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,04.56}{8,64}.100\%=25,9\%\\\%m_{Cu}=100\%-25,9\%=74,1\%\end{matrix}\right.\)

b) BTNT Fe, Cu: \(\left\{{}\begin{matrix}n_{Fe\left(NO_3\right)_3}=n_{Fe}=0,04\left(mol\right)\\n_{Cu\left(NO_3\right)_2}=n_{Cu}=0,1\left(mol\right)\end{matrix}\right.\)

=> m_muối = 0,04.242 + 0,1.188 = 28,48 (g)

BTNT N: \(n_{HNO_3}=3n_{Fe\left(NO_3\right)_3}+2n_{Cu\left(NO_3\right)_2}+n_{NO}+n_{NO_2}=0,48\left(mol\right)\)

=> \(V_{ddHNO_3}=\dfrac{0,48}{2}=0,24\left(l\right)=240\left(ml\right)\)

chỗ n_HNO3 em thấy hình như bị nhầm rồi ạ. em nghĩ là n_HNO3=n_NO2 + 3n_NO á