a)
\(2R+2aHCl-->2RCl_a+aH2\)
\(m_{HCl}=\frac{75.14,6}{100}=10,95\left(g\right)\)
\(n_{HCl}=\frac{10,95}{36,5}=0,3\left(mol\right)\)
\(n_R=\frac{1}{a}n_{HCl}=\frac{0,3}{a}\left(mol\right)\)
\(M_R=2,7:\frac{0,3}{a}=9a\)
\(a=3\Rightarrow R=27\left(Al\right)\)
Vậy R là Nhôm . Kí hiệu : Al
b) \(2Al+6HCl-->2AlCl3+3H2\)
\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\)
\(nHCl=0,2.1,5=0,3\left(mol\right)\)
Lập tỉ lệ
\(n_{Al}\left(\frac{0,4}{2}\right)>n_{HCl}\left(\frac{0,3}{6}\right)\)
\(\Rightarrow Al\) dư . Tính theo chất hết là HCl
\(n_{H2}=\frac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(V_{HCl}=0,15.22,4=3,36\left(l\right)\)
