C

\(n_{H_2}=\dfrac{5.376}{22,4}=0,24\left(mol\right)\)

PTHH : R + H₂SO₄ -> RSO₄ + H₂

           0,24                              0,24

\(M_R=\dfrac{9.6}{0,24}=40\left(\dfrac{g}{mol}\right)\)

Chọn C

\(R+2HCl\rightarrow RCl_2+H_2\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ n_R=n_{H_2}=0,1\left(mol\right)\\ M_R=\dfrac{6,5}{0,1}=65\left(\dfrac{g}{mol}\right)\\ \Rightarrow R\left(II\right):Kẽm\left(Zn=65\right)\)

R+ 2HCl →RCl₂ + H₂

0,1                    ←        0,1 mol 

n H₂ = 2,24:22,4=0,1 mol 

n R =6,5/M_R

=> 6,5/M_R =0,1 => M_R =65 

=> R là kẽm (Zn)

\(n_R=\dfrac{13}{M_R}\left(mol\right)\)

PTHH: R + H₂SO₄ --> RSO₄ + H₂

____\(\dfrac{13}{M_R}\)------------->\(\dfrac{13}{M_R}\)-->\(\dfrac{13}{M_R}\)

=> \(\dfrac{13}{M_R}\left(M_R+96\right)=32,2\)

=> M_R = 65(g/mol)

=> R là Zn

\(n_{H_2}=\dfrac{13}{65}=0,2\left(mol\right)\)

=> V_H2 = 0,2.22,4 = 4,48(l)

Bài 1 : 

$R + 2HCl \to RCl_2 + H_2$
n R = n H2 = 2,24/22,4 = 0,1(mol)

M R = 2,4/0,1 = 24(Mg) - Magie

Bài 2 : 

$2R + 6HCl \to 2RCl_3 + 3H_2$
n H2 = 3,36/22,4 = 0,15(mol)

n R = 2/3 n H2 = 0,1(mol)

M R = 2,7/0,1 = 27(Al) - Nhôm

\(a.R+2H_2O\rightarrow R\left(OH\right)_2+H_2\\ b.n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\\ n_R=n_{H_2}=0,5mol\\ M_R=\dfrac{20}{0,5}=40g/mol\)
Vậy kim loại R là Ca

a, PT: \(R+H_2SO_4\rightarrow RSO_4+H_2\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,4}{2}=0,2\left(l\right)\)

Theo ĐLBT KL, có: m_KL + m_H2SO4 = m _muối + m_H2

⇒ m _muối = 7,8 + 0,4.98 - 0,4.2 = 46,2 (g)

c, Gọi: n_R = x (mol) → n_Al = 2x (mol)

Theo PT: \(n_{H_2}=n_R+\dfrac{3}{2}n_{Al}=x+\dfrac{3}{2}.2x=0,4\left(mol\right)\Rightarrow x=0,1\left(mol\right)\)

⇒ n_R = 0,1 (mol)

n_Al = 0,1.2 = 0,2 (mol)

⇒ 0,1.M_R + 0,2.27 = 7,8 ⇒ M_R = 24 (g/mol)

Vậy: R là Mg.

\(R+2H_2O->R\left(OH\right)_2+H_2\\ n_R=n_{ROH}\\ \Rightarrow16,44:M_R=\dfrac{20,52}{M_R+17\cdot2}\\ M_R=137\left(Ba:barium\right)\)

\(n_R=\dfrac{16,44}{R}\left(mol\right);n_{R\left(OH\right)_2}=\dfrac{20,52}{R+\left(1+16\right).2}=\dfrac{20,52}{R+34}\left(mol\right)\\ R+H_2O\xrightarrow[]{}R\left(OH\right)_2+H_2\\ \Rightarrow n_R=n_{R\left(OH\right)_2}\\ \Leftrightarrow\dfrac{16,44}{R}=\dfrac{20,52}{R+34}\\ \Leftrightarrow16,44.\left(R+34\right)=R.20,52\\ \Leftrightarrow16,44R+558,96=20,52R \\ \Leftrightarrow558,96=20,52R-16,44R\\ \Leftrightarrow558,96=4,08R\\ \Leftrightarrow R=137\\\)

⇒R là Ba(Bari, 137)

\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

PTHH: R + 2HCl --> RCl₂ + H₂

        0,05<---------------0,05

=> \(M_R=\dfrac{1,2}{0,05}=24\left(g/mol\right)\)

=> R là Mg (Magie)

tysm!

nH2 = 1,2395/24,79 = 0,05 (mol)

PTHH: R + 2HCl -> RCl2 + H2

nR = 0,05 (mol)

M(R) = 2,8/0,05 = 56 (g/mol)

=> R là Fe

nH2 = 1,2395 : 24,79  = 0,05 (mol) 
pthh : R + 2HCl ---> RCl2  + H2 
           0,05 <-----------------0,05 (mol) 
=> M_R = 2,8 : 0,05 = 56 (g/mol ) 
=> R : Fe