\(n_{Al}=\dfrac{m}{M}=\dfrac{5,4}{27}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,2 0,3 ( mol )
\(m_{AlCl_3}=n_{AlCl_3}.M_{AlCl_3}=0,2.133,5=26,7g\)
\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)
2Al+6HCl->2AlCl3+3H2
0,2----0,6-----0,2------0,3
n Al=\(\dfrac{5,4}{27}\)=0,2 mol
=>m Alcl3=0,2.133,5=26,7g
=>VH2=0,3.22,4=6,72l