a) 

\(n_{H_2}=\dfrac{0,6}{2}=0,3\left(mol\right)\)

PTHH: 2M + 3H₂SO₄ --> M₂(SO₄)₃ + 3H₂

           0,2<----0,3<--------0,1<-------0,3

=> \(M_M=\dfrac{5,4}{0,2}=27\left(g/mol\right)\)

=> M là Al

b) \(C\%_{dd.H_2SO_4}=\dfrac{0,3.98}{395,2}.100\%=7,44\%\)

c) 

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

          0,2-->0,6

=> \(V_{dd.HCl}=\dfrac{0,6}{1}=0,6\left(l\right)=600\left(ml\right)\)

a) PTHH : 2M + 3H2SO4 --> M2(SO4)3 +3H2

n_M =5,4/M_M(mol)

n_H2=0,6/2=0,3(mol)

theo pthh ta có : n_M=2/3nH2=0,2(mol)

=> 5,4/M_M=0,2 => M_M =5,4/0,2=27(g/mol)=> M : Al

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

theo PTHH ta có : n_H2SO4=3/2n_Al=0,3(mol)

=>m_H2SO4=0,3.98=29,4(g)

=> C%dd_H2SO4=29,4/395,2 .100=7,44(%)

Bài 1:

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2\\ n_{Al}=\dfrac{2}{3}.0,3=0,2\left(mol\right)\\ \Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\\ \%m_{Al}=\dfrac{5,4}{26,82}.100\approx20,134\%\\\Rightarrow \%m_{Al_2O_3}\approx79,866\%\\ b,n_{Al_2O_3}=\dfrac{26,82-5,4}{102}=0,21\left(mol\right)\\ n_{HCl}=6.0,21+2.0,3=1,86\left(mol\right)\\ V_{ddHCl}=\dfrac{1,86}{2}=0,93\left(l\right)=930\left(ml\right)\\ m_{ddHCl}=930.1,12=1041,6\left(g\right)\\ n_{AlCl_3}=2.0,21+0,2=0,62\left(mol\right)\\ C\%_{ddAlCl_3}=\dfrac{0,62.133,5}{1041,6-0,3.2}.100\approx7,951\%\)

2)

a) Gọi KL và oxit của nó là M và MO

n_HCl = 4.0,25 = 1 (mol)

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: M + 2HCl --> MCl₂ + H₂

          0,3<-0,6<--------------0,3

            MO + 2HCl --> MCl₂ + H₂O

          0,2<---0,4

=> 0,3.M_M + 0,2.(M_M + 16) = 31,2

=> M_M = 56 (g/mol)

=> Kim loại là Sắt (Fe)

b) 

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,3.56}{31,2}.100\%=53,85\%\\\%m_{FeO}=\dfrac{0,2.72}{31,2}.100\%=46,15\%\end{matrix}\right.\)

Bài 2:

Đặt A là kim loại cần tìm

\(a,A+2HCl\rightarrow ACl_2+H_2\left(1\right)\\ AO+2HCl\rightarrow ACl_2+H_2O\left(2\right)\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ n_{HCl\left(ban.đầu\right)}=0,25.4=1\left(mol\right)\\ n_{HCl\left(1\right)}=2.n_{H_2}=0,6\left(mol\right)\\ n_{HCl\left(2\right)}=1-0,6=0,4\left(mol\right)\\ n_{AO}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ n_A=n_{H_2}=0,3\left(mol\right)\\ Vì:m_{hhđầu}=31,2\\ \Leftrightarrow m_A+m_{AO}=31,2\\ \Leftrightarrow0,3.M_A+0,2.\left(M_A+16\right)=31,2\\ \Leftrightarrow M_A=56\left(\dfrac{g}{mol}\right)\\ \Rightarrow A:Sắt\left(Fe=56\right)\\ b,\%m_{Fe}=\dfrac{0,3.56}{31,2}.100\approx53,846\%\\ \Rightarrow\%m_{FeO}\approx46,154\%\)

\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)

\(R+2HCl\rightarrow RCl_2+H_2\)

\(0.1........0.2................0.1\)

\(M_R=\dfrac{13.7}{0.1}=137\left(\dfrac{g}{mol}\right)\)

\(R:Ba\)

\(200\left(ml\right)=0.2\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0.2}{0.2}=1\left(M\right)\)

\(n_R=\dfrac{5,4}{R}\left(mol\right)\)

\(2R\left(\dfrac{5,4}{R}\right)+3H_2SO_4\left(\dfrac{8,1}{R}\right)\rightarrow R_2\left(SO_4\right)_3\left(\dfrac{2,7}{R}\right)+3H_2\left(\dfrac{8,1}{R}\right)\)

Theo PTHH \(n_{R_2\left(SO_4\right)_3}=\dfrac{2,7}{R}\left(mol\right)\)

\(\Rightarrow m_{R_2\left(SO_4\right)_3}=\dfrac{2,7.\left(2R+288\right)}{R}\left(g\right)\) \(\left(1\right)\)

\(n_{H_2}=\dfrac{8,1}{R}\left(mol\right)\)

\(\Rightarrow m_{H_2}=\dfrac{16,8}{R}\left(g\right)\)

\(m_{ddsau}=5,4+395,2-\dfrac{16,8}{R}\left(g\right)\)

Theo đề dung dịch muối sau phản ứng có nồng độ 8,55%

\(\Rightarrow8,55=\dfrac{\dfrac{2,7.\left(2R+288\right)}{R}}{5,4+395,2-\dfrac{16,8}{R}}.100\)

\(\Rightarrow R=27\left(Al\right)\)

Theo PTHH: \(n_{H_2SO_4}=\dfrac{8,1}{R}=\dfrac{8,1}{27}=0,3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=29,4\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}\left(bđ\right)=\dfrac{29,4}{395,2}.100=7,44\%\)

a) Gọi hóa trị của kim loại A là n(n>0,n∈Z)

\(n_A=\dfrac{5,4}{A}mol\\ n_{A_2\left(SO_4\right)_n}=\dfrac{342.10\%}{100\%.\left(2A+96n\right)}mol\\ 2A+nH_2SO_4\rightarrow A_2\left(SO_4\right)_n+nH_2\\ \Rightarrow n_A:2=n_{A_2\left(SO_4\right)_n}\\ \Leftrightarrow\dfrac{5,4}{A}:2=\dfrac{342.10\%}{100\%\left(2A+96n\right)}\\ \Leftrightarrow A=9n\)

Với n = 3 thì A = 27(TM)

Vậy kim loại A là Nhôm