Th1: R thuộc hóa trị I
\(2R+2HCl\rightarrow2RCl+H_2\)
Ta có: \(n_R=n_{RCl}\)
⇔ \(\dfrac{m}{2M_R}=\dfrac{m}{2M_{RCl}}\)
⇔ \(\dfrac{5,4}{2M_R}=\dfrac{26,7}{2\left(M_R+35,5\right)}\)
⇔ \(M_R=9\) (g/mol) (loại)
Th2: R thuộc hóa trị II
\(R+2HCl\rightarrow RCl_2+H_2\)
Ta có: \(n_R=n_{RCl_2}\)
⇔ \(\dfrac{m}{M_R}=\dfrac{m}{M_{RCl_2}}\)
⇔ \(\dfrac{5,4}{M_R}=\dfrac{26,7}{M_R+35,5.2}\)
⇔ \(M_R=18\) (g/mol) (loại)
Th3: R thuộc hóa trị III
\(2R+6HCl\rightarrow2RCl_3+3H_2\)
Ta có: \(n_R=n_{RCl_3}\)
⇔ \(\dfrac{m}{2M_R}=\dfrac{m}{2M_{RCl_3}}\)
⇔ \(\dfrac{5,4}{2M_R}=\dfrac{26,7}{2\left(M_R+35,5.3\right)}\)
⇔ \(M_R=27\) (g/mol) (nhận)
⇒ R là Nhôm (Al)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2-------0,6--------------------0,3(mol)
\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)
\(m_{ddHCl}=\dfrac{m_{HCl}.100}{C\%}=\dfrac{21,9.100}{73}=30\left(g\right)\)
\(V_{H_2}=n.22,4=0,3.22,4=6,72\left(l\right)\)