a)

Gọi n O2= a(mol) ; n N2 = b(mol)

Coi n X = 1(mol) ; M X = 14,8.2 = 29,6 (g/mol)

Ta có :

n X = a +  b = 1(mol)

m X = 32a + 28b = 29,6.1 = 29,6(gam)

Suy ra : a =0,4 ; b = 0,6

Ta có :

V O2 : V N2 = 0,4 : 0,6 = 2 : 3

b)

%m O2 = 0,4.32/29,6  .100% = 43,24%
%m N2 = 100%- 43,24% = 56,76%

\(GS:n_X=1\left(mol\right)\)

\(n_{O_2}=a\left(mol\right),n_{N_2}=b\left(mol\right)\)

\(\Leftrightarrow a+b=1\left(1\right)\)

\(M_X=14.8\cdot2=29.6\left(\dfrac{g}{mol}\right)\)

\(m_X=29.6\cdot1=29.6\left(g\right)\)

\(\Leftrightarrow32a+28b=29.6\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.4,b=0.6\)

\(V_{O_2}:V_{N_2}=0.6:0.4=3:2\)

\(\%m_{O_2}=\dfrac{0.4\cdot32}{29.6}\cdot100\%=43.24\%\)

\(\%m_{N_2}=56.76\%\)

\(\)

\(GS:n_A=1\left(mol\right)\)

\(n_{N_2}=a\left(mol\right),n_{H_2}=b\left(mol\right)\)

\(\Leftrightarrow a+b=1\left(1\right)\)

\(M_A=0.35\cdot28=9.8\left(\dfrac{g}{mol}\right)\)

\(m_A=9.8\cdot1=9.8\left(g\right)\)

\(\Leftrightarrow28a+2b=9.8\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.3,b=0.7\)

\(\%n_{N_2}=\dfrac{0.3}{1}\cdot100\%=30\%\)

\(\%n_{H_2}=70\%\)

\(\%m_{N_2}=\dfrac{0.3\cdot28}{9.6}\cdot100\%=87.5\%\)

\(\%m_{H_2}=12.5\%\)

\(\)

a) Gọi n N2 = a(mol) ; n H2 = b(mol)

Coi n A = 1(mol)

M A  = 0,35.28 = 9,8 (g/mol)

Ta có : 

a + b = 1

28a + 2b = 9,8.1

Suy ra a = 0,3 ; b = 0,7

%V N2 = 0,3/1 .100% = 30%

%V H2 = 0,7/1  .100% = 70%

b)

%m N2 = 0,3.28/9,8  . 100% = 85,71%
%m H2 = 100%- 85,71% = 14,29%

\(n_{SO_2}=\dfrac{4.48}{22.4}=0.2\left(mol\right)\)

\(\Rightarrow n_S=0.2\left(mol\right)\)

\(m_H=m_A-m_S=6.8-0.2\cdot32=0.4\left(g\right)\)

\(CT:H_xS_y\)

\(x:y=0.4:0.2=2:1\)

\(CTPT:H_2S\)

Khí A đốt cháy thu được nước và SO2 nên CTHH của A là $H_nS$

Bảo toàn nguyên tố với S : 

$n_{H_nS} = n_{SO_2} = \dfrac{4,48}{22,4} = 0,2(mol)$

$\Rightarrow m_A = n + 32 = \dfrac{6,8}{0,2} = 34$

$\Rightarrow n = 2$

Vậy CTPT của A là $H_2S$

$M_2SO_3 + 2HCl \to 2MCl + SO_2 + H_2O$

$MHCO_3 + HCl \to MCl + CO_2 + H_2O$
$MHSO_3 + HCl \to MCl + SO_2 + H_2O$

Theo PTHH : 

n X = n khí = 1,792/22,4 = 0,08(mol)

=> MX = 8,16/0,08 = 102

Ta có : 

M + 61 < 2M + 81 < 2M + 80  

<=> M + 61 < 102 < 2M + 80

<=> 11 < M < 41

Vậy M có thể là Natri hoặc Kali

\(n_{SO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(T=\dfrac{0.25}{0.3}=0.83\)

=> Tạo ra CaSO₃

\(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)

\(0.25..........0.25...........0.25\)

\(m_{CaSO_3}=0.25\cdot120=30\left(g\right)\)

\(m_{Ca\left(OH\right)_2\left(dư\right)}=\left(0.3-0.25\right)\cdot74=3.7\left(g\right)\)

PTHH: \(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3\downarrow+H_2O\)

Ta có: \(n_{SO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)=n_{BaSO_3}\)

\(\Rightarrow m_{BaSO_3}=0,15\cdot217=32,55\left(g\right)\)

        PTHH: \(MgO+2HCl\rightarrow MgCl_2+H_2O\)

                        a______2a______a               (mol)

                     \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

                         b_____2b______b                 (mol)

Ta lập được HPT: \(\left\{{}\begin{matrix}40a+80b=20\\2a+2b=0,2\cdot3=0,6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgO}=\dfrac{0,1\cdot40}{20}\cdot100\%=20\%\\\%m_{CuO}=80\%\\m_{MgCl_2}=0,1\cdot95=9,5\left(g\right)\\m_{CuCl_2}=0,2\cdot135=27\left(g\right)\end{matrix}\right.\)

a)

$MgO + 2HCl \to MgCl_2 + H_2O$
$CuO + 2HCl \to CuCl_2 + H_2O$

b)

Gọi n MgO = a(mol) ; n CuO =b(mol)

=> 40a + 80b = 20(1)

Theo PTHH : 

n HCl = 2a + 2b = 0,2.3 =0,6(2)

Từ (1)(2) suy ra a = 0,1 ; b = 0,2

%m MgO = 0,1.40/20  .100% = 20%
%m CuO = 100% -20% = 80%

c)

n MgCl2 = a = 0,1(mol)

n CuCl2 = b = 0,2(mol)

=> m = 0,1.95 + 0,2.135 = 36,5 gam

\(a.\)

\(BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(K_2O+H_2O\rightarrow2KOH\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

\(b.\)

\(SO_2+H_2O⇌H_2SO_3\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

\(CO_2+H_2O⇌H_2CO_3\)

\(c.\)

\(BaO+2HCl\rightarrow BaCl_2+H_2O\)

\(FeO+2HCl\rightarrow FeCl_2+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(K_2O+2HCl\rightarrow2KCl+H_2O\)

\(CaO+2HCl\rightarrow CaCl_2+H_2O\)

\(d.\)

\(Ba\left(OH\right)_2+SO_2\rightarrow BaSO_3+H_2O\)

\(Ba\left(OH\right)_2+2SO_2\rightarrow Ba\left(HSO_3\right)_2\)

\(Ba\left(OH\right)_2+SO_3\rightarrow BaSO_4+H_2O\)

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

\(Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\)

\(n_{MgO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)

\(m_{hh}=40a+160b=24\left(g\right)\left(1\right)\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

\(n_{HCl}=2a+6b=1\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.1\)

\(m_{MgO}=0.2\cdot40=8\left(g\right)\)

\(m_{Fe_2O_3}=16\left(g\right)\)

Bài 2 : 

\(n_{CO_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

\(..............0.25.......0.25\)

\(m_{CaCO_3}=0.25\cdot100=25\left(g\right)\)

Bài 1 : 

\(n_{Al_2O_3}=a\left(mol\right),n_{MgO}=b\left(mol\right)\)

\(m_{hh}=102a+40b=26.4\left(g\right)\left(1\right)\)

\(Al_2O_3+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

\(n_{H_2SO_4}=3a+b=0.75\left(mol\right)\left(2\right)\)

\(\left(1\right),\left(2\right):a=0.2,b=0.15\)

\(m_{Al_2O_3}=0.2\cdot102=20.4\left(g\right)\)

\(m_{MgO}=0.15\cdot40=6\left(g\right)\)

\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.2}{0.75}=0.266\left(M\right)\)

\(C_{M_{MgSO_4}}=\dfrac{0.15}{0.75}=0.2\left(M\right)\)