a) \(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)

PTHH: 2M + 6HCl → 2MCl₃ + 3H₂

Mol:     0,02    0,06         0,02      0,03

\(M_M=\dfrac{0,54}{0,02}=27\left(g/mol\right)\)

  ⇒ M là nhôm (Al)

\(C\%_{ddHCl}=\dfrac{0,06.36,5.100\%}{500}=0,438\%\)

c) m_{dd sau pứ} = 0,54 + 500 - 0,03.2 = 500,48 (g)

\(C\%_{ddAlCl_3}=\dfrac{0,02.133,5.100\%}{500,48}=0,53\%\)

n_Mg = 0,1(mol)

PTHH: Mg + 2HCl --> MgCl₂ +H₂

n_Mg = n_MgCl2= n_H2 = 0,1(mol)

=> m_muối = 9,5(g) 

_VH2 = 2,24(l)

b) CM_HCl = 0,2/0,1=2(M)

a) Gọi kim loại cần tìm là R

\(n_R=\dfrac{7,56}{M_R}\left(mol\right)\)

PTHH: 2R + 2nHCl --> 2RCl_n + nH₂

         \(\dfrac{7,56}{M_R}\)------------>\(\dfrac{7,56}{M_R}\)

=> \(M_{RCl_n}=M_R+35,5n=\dfrac{37,38}{\dfrac{7,56}{M_R}}\)

=> \(M_R=9n\left(g/mol\right)\)

Xét n = 1 => M_R = 9(Loại)

Xét n = 2 => M_R = 18 (Loại)

Xét n = 3 => M_R = 27(g/mol) => R là Al (Nhôm)

b) 

\(n_{Al}=\dfrac{7,56}{27}=0,28\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,28-->0,84--->0,28--->0,42

=> \(V_{H_2}=0,42.22,4=9,408\left(l\right)\)

\(m_{HCl}=0,84.36,5=30,66\left(g\right)\)

=> \(m_{ddHCl}=\dfrac{30,66.100}{12}=255,5\left(g\right)\)

c) m_{dd sau pư} = 7,56 + 255,5 - 0,42.2 = 262,22 (g)

=> \(C\%_{AlCl_3}=\dfrac{37,38}{262,22}.100\%=14,255\%\)

\(a,n_{CO_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:M_2CO_3+2HCl\to 2MCl+H_2O+CO_2\uparrow\\ \Rightarrow n_{M_2CO_3}=n_{CO_2}=0,15(mol)\\ \Rightarrow M_{M_2CO_3}=\dfrac{15,9}{0,15}=106(g/mol)\\ \Rightarrow M_{M}=\dfrac{106-12-16.3}{2}=23(g/mol)\)

Vậy M là natri (Na)

\(b,n_{HCl}=2n_{CO_2}=0,3(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,3}{0,75}=0,4(l)\\ X:NaCl\\ n_{NaCl}=n_{HCl}=0,3(mol)\\ \Rightarrow C_{M_{NaCl}}=\dfrac{0,3}{0,4}=0,75M\)

mik đã giải chi tiết lắm rồi nếu ko hiểu bn hỏi lại nhé !

\(\text{Ta có }n_M=\dfrac{10,8}{M_M}\left(mol\right);n_{N_2O}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:8M+30HNO_3\rightarrow8M\left(NO_3\right)_3+3N_2O+15H_2O\\ \Rightarrow n_M=\dfrac{8}{3}n_{N_2O}=0,4\left(mol\right)\\ \Rightarrow\dfrac{10,8}{M_M}=0,4\\ \Rightarrow M_M=27\)

Vậy M là nhôm (Al)

CTHH: A_xO_y

\(n_{H_2}=\dfrac{1,344}{22,4}=0,06\left(mol\right)\)

PTHH: A_xO_y + yH₂ --t^o--> xA + yH₂O

          \(\dfrac{0,06}{y}\)<--0,06---->\(\dfrac{0,06x}{y}\)

            2A + 2nHCl --> 2ACl_n + nH₂

        \(\dfrac{0,06x}{y}\)---------------->\(\dfrac{0,03xn}{y}\)

=> \(\dfrac{0,03xn}{y}=\dfrac{1,008}{22,4}=0,045\left(mol\right)\)

=> \(\dfrac{y}{x}=\dfrac{2}{3}n\)

\(M_{A_xO_y}=\dfrac{3,48}{\dfrac{0,06}{y}}=58y\left(g/mol\right)\)

=> \(x.M_A=42y\)

=> \(M_A=\dfrac{42y}{x}=28n\left(g/mol\right)\)

Xét n = 2 thỏa mãn => M_A = 56 (g/mol)

=> A là Fe

\(\dfrac{x}{y}=\dfrac{3}{2n}=\dfrac{3}{4}\) => CTHH: Fe₃O₄