Đáp án:
a) $2Al+6HCl→2AlCl_3+3H_2$
$Al_2O_3+6HCl→2AlCl_3+3H_2O$
b) %$m_{Al}$ $=$ $34,62$%
%$m_{Al_2O_3}=65,38$%
Giải thích các bước giải:
a) PTHH:
$2Al+6HCl→2AlCl_3+3H_2$
$Al_2O_3+6HCl→2AlCl_3+3H_2O$
b) $n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)$
$n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}×0,3=0,2(mol)$
→ $m_{Al}=0,2×27=5,4(g)$
→ $m_{Al_2O_3}=m_{hh}-m_{Al}=15,6-5,4=10,2(g)$
%$m_{Al}=\dfrac{5,4}{15,6}×100$% $=$ $34,62$%
→ %$m_{Al_2O_3}=65,38$%
