a) pt: $2Al+6HCl\rightarrow 2AlCl_{3}+3H_{2}$ 0.4 $ \leftarrow $ 0.6 (mol)$Al_{2}O_{3}+6HCl \rightarrow 2AlCl_{3}+3H_{2}O$ $x$ (mol)$=>$ Ta được : $m_{Al}+m_{Al_{2}O_{3}}=21<=>m_{Al_{2}O_{3}}=21-0,4.27=10,2(g)$
a) pt : $2Al+6HCl\rightarrow 2AlCl_{3}+3H_{2}$ 0.4 $ \leftarrow $ 0.6 (mol)$Al_{2}O_{3}+6HCl \rightarrow 2AlCl_{3}+3H_{2}O$ $x$ (mol)$=>$ Ta được : $m_{Al}+m_{Al_{2}O_{3}}=21<=>m_{Al_{2}O_{3}}=21-0,4.27=10,2(g)$
a) pt: $2Al+6HCl\rightarrow 2AlCl_{3}+3H_{2}$ 0.4 $ \leftarrow $ 0.6 (mol)$Al_{2}O_{3}+6HCl \rightarrow 2AlCl_{3}+3H_{2}O$ $x$ (mol)$=>$ Ta được : $m_{Al}+m_{Al_{2}O_{3}}=21<=>m_{Al_{2}O_{3}}=21-0,4.27=10,2(g)$