$a,PTPƯ:2Al+6HCl\xrightarrow{} 2AlCl_3+3H_2↑$
$n_{Al}=\dfrac{2,7}{27}=0,1mol.$
$Theo$ $pt:$ $n_{H_2}=\dfrac{3}{2}n_{Al}=0,15mol.$
$⇒V_{H_2}=0,15.22,4=3,36l.$
$b,Theo$ $pt:$ $n_{HCl}=3n_{Al}=0,3mol.$
$⇒m_{ddHCl}=\dfrac{0,3.36,5}{7,3\%}=150g.$
$Theo$ $pt:$ $n_{AlCl_3}=n_{Al}=0,1mol.$
$⇒C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+150-(0,1.2)}.100\%=8,75\%$
chúc bạn học tốt!
