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

PTHH: 2M + H₂O → 2MOH + H₂

Mol:     0,1                   0,1      0,05

\(M_M=\dfrac{3,9}{0,1}=39\left(g/mol\right)\)

⇒ M là kali (K)

b, \(m_{KOH}=0,1.56=5,6\left(g\right)\)

\(n_{H_2}=\dfrac{0,336}{22,4}=0,015mol\)

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

0,01    0,03      0,01       0,015

\(M\)à \(n_R=\dfrac{m_R}{M_R}=\dfrac{0,27}{M_R}=0,01\)\(\Rightarrow M_R=27\left(đvC\right)\)

R là kim loại Al.

\(m_{RCl_3}=0,01\cdot133,5=1,335\left(g\right)\)

\(m_{H_2}=0,015\cdot2=0,03\left(g\right)\)

BTKL: \(m_{ddsau}=0,27+300-0,03=300,24\left(g\right)\)

\(C\%=\dfrac{1,335}{300,24}\cdot100\%=44,46\%\)

PTHH: 2R + 6HCl ---> 2RCl₃ + 3H₂ (1)

Ta có: \(n_{H_2}=\dfrac{0,336}{22,4}=0,015\left(mol\right)\)

Theo PT₍₁₎: \(n_R=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,015=0,01\left(mol\right)\)

=> \(M_R=\dfrac{0,27}{0,01}=27\left(g\right)\)

Vậy R là nhôm (Al)

PTHH: 2Al + 6HCl ---> 2AlCl₃ + 3H₂ (2)

Theo PT₍₂₎: \(n_{AlCl_3}=n_{Al}=0,01\left(mol\right)\)

=> \(m_{AlCl_3}=0,01.133,5=1,335\left(g\right)\)

Ta có: \(m_{dd_{AlCl_3}}=0,27+300-0,015.2=300,24\left(g\right)\)

=> \(C_{\%_{AlCl_3}}=\dfrac{1,335}{300,24}.100\%=0,44\%\)