Mọi người giúp mik với ajajjajak

$\rm n_{HCl}=\dfrac{73.20}{100.36,5}=0,4(mol)$

$\rm PTHH:2Al+6HCl \to 2AlCl_3+3H_2$

$\rm a.n_{H_2}=\dfrac{1}{2}n_{HCl}=0,2(mol)$

$\rm V=0,2.24,79=4,958(l)$

$\rm b.n_{Al}=n_{AlCl_3}=\dfrac{1}{3}n_{HCl}=\dfrac{2}{15}(mol)$

$\rm m_{Al}=\dfrac{2}{15}.133,5=17,8(g)$

$\rm m_{dd sau}=73+3,6-0,2.2=76,2(g)$

$\rm C\%=\dfrac{\dfrac{2}{15}.133,5}{76,2}.100\%=23,36\%$