Đáp án:
a) $\%m_{Al}=34,61\%;\ \%m_{Al_2O_3}= 65,39\%$
b) $V_{dd\ HCl}=1,2\ l$
c) $m_{AlCl_3}=26,7\ g$
Giải thích các bước giải:
Ta có: $n_{H_2}=\dfrac V{22,4} = \dfrac{3,36}{22,4}=0,15\ mol$
a/
PTHH: $2Al+6HCl \to 2AlCl_3+3H_2$ (1)
$Al_2O_3+6HCl \to 2AlCl_3+3H_2O$ (2)
Theo (1) $n_{Al}= \dfrac 23 n_{H_2}=\dfrac 23 0,15=0,1\ mol$
Vậy: $\%m_{Al}=\dfrac{0,1.27}{7,8}=34,61\%$
$\to \%m_{Al_2O_3}=100 - 34,61 = 65,39\%$
b/
$n_{Al_2O_3}=\dfrac{7,8 - 0,1.27}{102}=0,05\ mol$
Theo (1), (2): $n_{HCl}=3n_{Al}+6n_{Al_2O_3}=3.0,1+6.0,05=0,6\ mol$
$ \to V_{dd\ HCl}=\dfrac{0,6}{0,5}=1,2\ l$
c/ Theo (1), (2): $n_{AlCl_3}=n_{Al}+2n_{Al_2O_3}=0,1+0,05.2=0,2\ mol$
$\to m_{AlCl_3}=0,2.133,5=26,7\ g$