Giải thích các bước giải:
\(\begin{array}{l}
2.\\
a)Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}O\\
b)\\
{n_{Fe}} = 0,4mol\\
{n_{{H_2}S{O_4}}} = 0,25mol\\
\to {n_{{H_2}S{O_4}}} < {n_{Fe}} \to {n_{Fe}}dư\\
\to {n_{Fe}} = {n_{{H_2}S{O_4}}} = 0,25mol\\
\to {n_{Fe}}dư = 0,15mol\\
\to {m_{Fe}}dư = 8,4mol\\
c)\\
{n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,25mol\\
\to {V_{{H_2}}} = 5,6l\\
d)\\
{n_{FeS{O_4}}} = {n_{{H_2}S{O_4}}} = 0,25mol\\
\to {m_{FeS{O_4}}} = 38g\\
3.\\
a)A{l_2}{O_3} + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}O\\
{n_{A{l_2}{O_3}}} = 0,6mol\\
{n_{{H_2}S{O_4}}} = 0,6mol\\
b)
\end{array}\)
Không có chất dư
\(\begin{array}{l}
c)\\
{n_{A{l_2}{{(S{O_4})}_3}}} = {n_{A{l_2}{O_3}}} = 0,6mol\\
\to {m_{A{l_2}{{(S{O_4})}_3}}} = 205,2g
\end{array}\)
