Giải thích các bước giải:
Câu 5:
\(\begin{array}{l}
3Fe + 2{O_2} \to F{e_3}{O_4}\\
{n_{Fe}} = 0,3mol\\
{n_{{O_2}}} = \dfrac{2}{3}{n_{Fe}} = 0,2mol \to {V_{{O_2}}} = 4,48l
\end{array}\)
Câu 6:
\(\begin{array}{l}
2Al + 6HCl \to 2AlC{l_3} + 3{H_2}\\
{n_{Al}} = 0,15mol\\
{n_{{H_2}}} = \dfrac{3}{2}{n_{Al}} = 0,225mol \to {V_{{H_2}}} = 5,04l\\
{n_{AlC{l_3}}} = {n_{Al}} = 0,15mol \to {m_{AlC{l_3}}} = 20,025g\\
{m_{{\rm{dd}}}} = 104,5g \to {m_{{\rm{dd}}HCl}}dư= 104,5 - 20,025 = 84,475g\\
{n_{HCl}} = 3{n_{Al}} = 0,45mol \to {m_{HCl}} = 16,425g\\
m = {m_{{\rm{dd}}HCl}}dư+ {m_{HCl}} = 100,9g
\end{array}\)
