$n_{Al}=\frac{5,4}{27}=0,2~(mol)$
$2Al+6HCl→2AlCl_3+3H_2↑$
$0,2$ $0,6$ $0,2$ $0,3$ $(mol)$
$a./$
$m_{ct~HCl}=0,6.(35,5+1)=21,9~(g)$
$m_{dd~HCl}=\frac{21,9. 100\ \%}{18,25\ \%}=120~(g)$
$b./$
$m_{ct~AlCl_3}=0,2.(27+106,5)=26,7~(g)$
$m_{dd~AlCl_3}=5,4+120-(0,3.2)=124,8~(g)$
$C\ \%_{AlCl_3}=\frac{26,7. 100\ \%}{124,8}≈21,394\ \%$
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