a, \(2M+3Cl_2\underrightarrow{^{to}}MCl_3\)
\(n_M=n_{MCl3}\Leftrightarrow\frac{10,8}{M}=\frac{53,4}{M+106,5}\)
\(\Leftrightarrow10,8M+1150,2=53,4M\)
\(\Leftrightarrow42,6M=1150,2\)
\(\Leftrightarrow M=27\left(\frac{g}{mol}\right)\)
Vậy M là Nhôm (Al)
b, \(2Al+3Cl_2\underrightarrow{^{to}}2AlCl_3\)
\(n_{Al}=\frac{10,8}{27}=0,4\left(mol\right)\)
\(n_{Cl2}=\frac{3}{2}n_{Al}=0,6\left(mol\right)\)
\(MnO_2+4HCl\rightarrow MnO_2+Cl_2+H_2O\)
\(n_{Cl2}=0,6\left(mol\right)\)
Mà \(H=80\%\rightarrow n_{Cl2}=0,6.80\%=0,75\left(mol\right)\)
\(m_{MnO2}=0,75.87=65,25\left(g\right)\)
\(m_{HCl}=4n_{Cl2}.36,5=109,5\left(g\right)\)
\(\rightarrow m_{dd_{HCl}}=\frac{109,5}{37}.100\%=295,95\left(g\right)\)
\(\rightarrow V_{HCl}=\frac{295,95}{1,19}=258,7\left(l\right)\)
