a.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,2                                         0,3    ( mol )

\(V_{H_2}=0,3.22,4=6,72l\)

b.\(XO+H_2\rightarrow\left(t^o\right)X+H_2O\)

\(n_X=\dfrac{19,5}{M_X}\) mol

\(n_{H_2}=n_X=0,3mol\)

\(\Rightarrow\dfrac{19,5}{M_X}=0,3\)

\(M_X=65\) ( g/mol )

=> X là kẽm ( Zn )

a, n_Al = \(\dfrac{5,4}{27}=0,2\left(mol\right)\)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂

           0,2      0,6             0,2      0,3

V_H2 = 0,3.22,4 = 6,72 (l)

b, PTHH: RO + H₂ ---t^o---> R + H₂O

                          0,3             0,3

=> M_R = \(\dfrac{19,5}{0,3}=65\left(\dfrac{g}{mol}\right)\)

=> R là Zn

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(2X+O_2\underrightarrow{t^o}2XO\)

\(\dfrac{13}{X}\)     0,1

\(\Rightarrow\dfrac{13}{X}=0,1\cdot2\Rightarrow X=65\)

Vậy X là kẽm Zn.

\(m_{ZnO}=0,2\cdot81=1,62g\)

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: 2R + O₂ --to--> 2RO

          0,2   0,.1

=> \(M_R=\dfrac{13}{0,2}=65\left(\dfrac{g}{mol}\right)\)

=> R: Zn

Bài 1 : 

Coi $m_{O_2} = 32(gam) \Rightarrow m_M = \dfrac{32}{40\%} = 80(gam)$

$n_{O_2} = 1(mol)$
$2M + O_2 \xrightarrow{t^o} 2MO$
$n_M = 2n_{O_2} = 2(mol)$
$M_M = \dfrac{80}{2} = 40$
Vậy M là kim loại Canxi

CTHH oxit là CaO

TD4 :

Đặt : CTHH là : \(H_xP_y\)

\(M_A=1.0625\cdot32=34\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow xH+yP=34\)

\(\Rightarrow x+31y=34\)

Chỉ có duy nhất một cặp nghiệm thỏa mãn : 

\(x=3,y=1\)

\(CT:PH_3\)

TD3 : 

\(n_{CO_2}=\dfrac{4.4}{44}=0.1\left(mol\right)\Rightarrow n_C=0.1\left(mol\right)\)

\(n_{H_2O}=\dfrac{2.7}{18}=0.15\left(mol\right)\Rightarrow n_H=2\cdot0.15=0.3\left(mol\right)\)

\(m_O=m_A-m_C-m_H=2.3-0.1\cdot12-0.3=0.8\left(g\right)\)

\(\Rightarrow n_O=\dfrac{0.8}{16}=0.05\left(mol\right)\)

Ta có : 

\(n_C:n_H:n_O=0.1:0.3:0.05=2:6:1\)

CTHH của A là : \(C_2H_6O\)

Bài 1:

\(n_M=\dfrac{16}{M_M}\left(mol\right)\)

PTHH: 2M + O₂ --t^o--> 2MO

         \(\dfrac{16}{M_M}\)---------->\(\dfrac{16}{M_M}\)

=> \(\dfrac{16}{M_M}\left(M_M+16\right)=20\)

=> M_M = 64 (g/mol)

=> M là Cu

Bài 2:

\(n_R=\dfrac{16,2}{M_R}\left(mol\right)\)

PTHH: 2R + 3Cl₂ --t^o--> 2RCl₃

          \(\dfrac{16,2}{M_R}\)------------>\(\dfrac{16,2}{M_R}\)

=> \(\dfrac{16,2}{M_R}\left(M_R+106,5\right)=80,1\)

=> M_R = 27 (g/mol)

=> R là Al

 1 
ADDDLBTKL ta có
\(m_{O_2}=m_{MO}-m_M\\ m_{O_2}=20-16=4g\\ n_{O_2}=\dfrac{4}{32}=0,125\left(mol\right)\\ pthh:2M+O_2\underrightarrow{t^o}2MO\) 
            0,25   0,125 
\(M_M=\dfrac{16}{0,25}=64\left(\dfrac{g}{mol}\right)\) 
=> M là Cu 
2 
ADĐLBTKL ta có 
\(m_{Cl_2}=m_{RCl_3}-m_R\\ m_{Cl_2}=80,1-16,2=63,9g\\ n_{Cl_2}=\dfrac{63,9}{71}=0,9\left(mol\right)\\ pthh:2R+3Cl_2\underrightarrow{t^o}2RCl_3\) 
            0,6   0,9 
\(M_R=\dfrac{16,2}{0,6}=27\left(\dfrac{g}{mol}\right)\) 
=> R là Al

\(1 ) 2M+O_2\rightarrow 2MO n_M=n_{MO}\Leftrightarrow \dfrac{16}{M_M}=\dfrac{20}{m_M+16} \Rightarrow m_m = 64(g/mol) \rightarrow M : Cu \)

\(2) 2R+3Cl_2\rightarrow 2RCl_3 n_R=nn_{RCl_3}\Leftrightarrow \dfrac{16,2}{M_R}=\dfrac{80,1}{M_R+35,5.3}\Rightarrow M_R = 27(g/mol)\rightarrow R:Al \)

PT: \(2R+O_2\underrightarrow{t^o}2RO\)

Ta có: \(n_R=\dfrac{13}{M_R}\left(mol\right)\), \(n_{RO}=\dfrac{16,2}{M_R+16}\left(mol\right)\)

Theo PT: \(n_R=n_{RO}\Rightarrow\dfrac{13}{M_R}=\dfrac{16,2}{M_R+16}\Rightarrow M_R=65\left(g/mol\right)\)

→ R là Kẽm (Zn).

2Al+3H2SO4->Al2(SO4)3+3H2

0,2-----------------------------------0,3

n Al=0,2 mol

=>VH2=0,3.22,4=6,72l

b)

XO+H2-to>X+H2O

0,3-------------0,3

=>0,3=\(\dfrac{19,5}{X}\)

=>X là Zn( kẽm)

a.\(n_{Al}=\dfrac{5,4}{27}=0,2mol\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

0,2                                    0,3  ( mol )

\(V_{H_2}=0,3.22,4=6,72l\)

b.\(n_X=\dfrac{19,5}{M_X}\)

\(XO+H_2\rightarrow\left(t^o\right)X+H_2O\)

          \(\dfrac{19,5}{M_X}\)        \(\dfrac{19,5}{M_X}\)         ( mol )

Ta có:

\(\dfrac{19,5}{M_X}=0,3\)

\(\Leftrightarrow M_X=65\)

=> X là kẽm (Zn)

\(a,n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\\ PTHH:2Al+6HCl\rightarrow AlCl_3+3H_2\uparrow\\ Theo.pt:n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ V_{H_2}=0,3.22,4=6,72\left(l\right)\\ b,PTHH:RO+H_2\underrightarrow{t^o}R+H_2O\\ Mol:0,3\leftarrow0,3\rightarrow0,3\\ M_R=\dfrac{19,5}{0,3}=65\left(\dfrac{g}{mol}\right)\\ \Rightarrow R.là.Zn\)

Đặt kim loại hóa trị II là A.

=> Oxit: AO

\(AO+2HCl\rightarrow ACl_2+H_2O\\ n_{AO}=n_A=\dfrac{1,9-0,8}{35,5.2-16}=0,02\left(mol\right)\\ M_{AO}=\dfrac{0,8}{0,02}=40\left(\dfrac{g}{mol}\right)\\ Mà:M_{AO}=M_A+16\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_A+16=40\\ \Leftrightarrow M_A=24\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Magie\left(Mg=24\right)\\ \Rightarrow Oxit:MgO\)

Gọi tên oxit: Magie oxit.