$n_{Al}=6,7/27=0,248mol$
$a.2Al+6HCl\to 2AlCl_3+3H_2↑$
b.Theo pt :
$n_{HCl}=3.n_{Al}=3.0,248=0,744mol$
$⇒m_{HCl}=0,744.36,5=27,156g$
$⇒m_{ddHCl}=\dfrac{27,156}{10\%}=271,56g$
c.Theo pt :
$n_{H_2}=3/2.n_{Al}=3/2.0,248=0,372mol$
$n_{AlCl_3}=n_{Al}=0,248mol$
$⇒m_{AlCl_3}=0,248.133,5=33,108g$
$m_{ddspu}=6,7+271,56-0,372.2=277,516g$
$⇒C\%_{AlCl_3}=\dfrac{33,108}{277,516}.100\%=27,12\%$