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

a. PTHH: R + H₂SO₄ ---> RSO₄ + H₂ (1)

Ta có: \(n_{H_2}=\dfrac{3,584}{22,4}=0,16\left(mol\right)\)

Theo PT₍₁₎: \(n_R=n_{H_2}=0,16\left(mol\right)\)

=> \(M_R=\dfrac{3,84}{0,16}=24\left(g\right)\)

Vậy R là magie (Mg)

b. PTHH:

Mg + HCl ---> MgCl₂ + H₂ (2)

Theo PT₍₂₎: \(n_{H_2}=n_{Mg}=0,16\left(mol\right)\)

=> \(V_{H_2}=0,16.22,4=3,584\left(lít\right)\)

(Do câu b đề ko rõ lắm nên mik làm như vậy, nếu sai bn bình luận nhé.)

\(PTHH:2R+6HCl\rightarrow2RCl_3+3H_2\)

\(TheoPTHH:n_R=n_M=\dfrac{10,8}{R}=\dfrac{53,4}{R+35,5.3}\)

\(\Rightarrow R=27\)

=> Kim loại đó là Nhôm

b, \(TheoPTHH:n_{HCl}=3n_R=1,5mol\)

\(\Rightarrow V_{HCl}=3l\)

Theo PTHH : \(n_{H2}=\dfrac{3}{2}n_{Al}=0,75mol\)

\(\Rightarrow V=n.22,4=16,8l\)

\(2M+6HCl\rightarrow2MCl_3+3H_2\)

\(2M...........2\cdot\left(M+106.5\right)\)

\(10.8..................53.4\)

\(53.4\cdot2M=10.8\cdot\cdot2\left(M+106.5\right)\)

\(\Rightarrow M=27\)

\(M:Nhôm\)

\(n_{Al}=\dfrac{13.5}{27}=0.5\left(mol\right)\)

\(V_{H_2}=0.5\cdot\dfrac{3}{2}\cdot22.4=16.8\left(l\right)\)

\(V_{dd_{HCl}}=\dfrac{0.5\cdot6}{2\cdot0.5}=3\left(l\right)\)

Gọi hóa trị của R là n

PTHH : \(2R+nH_2SO_4-->R_2\left(SO_4\right)n+nH_2\)

Theo pthh : \(n_{R2\left(SO4\right)n}=\dfrac{1}{2}n_R\)

\(\Rightarrow\dfrac{34,2}{2M_R+96n}=\dfrac{1}{2}.\dfrac{5,4}{M_R}\)

\(\Rightarrow M_R=9n\)

Ta có bảng sau :

Vậy R là kim loại Al

\(n_{H2\left(dktc\right)}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

a) Pt : \(2R+3H_2SO_4\rightarrow R_2\left(SO_4\right)_3+3H_2|\)

            2            3                    1              3

           0,2                                               0,3

\(n_R=\dfrac{0,3.2}{3}=0,2\left(mol\right)\)

⇒ \(M_R=\dfrac{5,4}{0,2}=27\left(dvc\right)\)

Vậy kim loại R là nhôm

b) \(2Al+6H_2SO_{4\left(đặc,nóng\right)}\rightarrow Al_2\left(SO_4\right)_3+3SO_2+6H_2O|\)

     2              6                                 1               3          6

   0,2                                                                0,3

\(n_{SO2}=\dfrac{0,3.3}{2}=0,3\left(mol\right)\)

\(V_{SO2\left(dktc\right)}=0,3.22,4=6,72\left(l\right)\)

 Chúc bạn học tốt 

a) PTHH: \(2R+3H_2SO_4\rightarrow R_2\left(SO_4\right)_3+3H_2\uparrow\)

Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\) \(\Rightarrow n_R=0,2\left(mol\right)\)

\(\Rightarrow M_R=\dfrac{5,4}{0,2}=27\) \(\Rightarrow\) R là Nhôm (Al)

b) PTHH: \(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

Theo PTHH: \(n_{SO_2}=0,3\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,3\cdot22,4=6,72\left(l\right)\) 

\(n_{H_2}=\dfrac{0.336}{22.4}=0.015\left(mol\right)\)

\(M+2HCl\rightarrow MCl_2+H_2\)

\(0.015........................0.015\)

\(M_M=\dfrac{0.6}{0.015}=40\left(\dfrac{g}{mol}\right)\)

\(M:Canxi\left(Ca\right)\)

\(n_{H_2SO_4}=0.1\cdot0.8=0.08\left(mol\right)\)

\(M+H_2SO_4\rightarrow MSO_4+H_2\)

\(0.08.....0.08\)

\(M_M=\dfrac{4.48}{0.08}=56\left(\dfrac{g}{mol}\right)\)

\(M:Sắt\left(Fe\right)\)

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

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)