\(n_{Al}=\frac{1,08}{27}=0,04mol\)
\(2Al+3Cl_2\rightarrow^{t^o}2AlCl_3\)
a) \(n_{Cl_2}=\frac{3}{2}.0,04=0,06mol\)
\(V_{Cl_2}=0,06.22,4=1,344l\)
b) Cách 1: \(m_{AlCl_3}=m_{Al}+m_{Cl_2}=1,08+71.0,06=5,34g\)
Cách 2: \(n_{AlCl_3}=n_{Al}=0,04mol\)
\(m_{AlCl_3}=0,04.133,5=5,34g\)