Em tham khảo nha:
\(\begin{array}{l}
3)\\
a)\\
{n_{Fe}} = \dfrac{{28}}{{56}} = 0,5\,mol\\
{n_{Cu}} = \dfrac{{64}}{{64}} = 1\,mol\\
{n_{Al}} = \dfrac{{5,4}}{{27}} = 0,2\,mol\\
b)\\
{V_{{O_2}}} = 0,125 \times 22,4 = 2,8l\\
{V_{C{O_2}}} = 0,25 \times 22,4 = 5,6l\\
c)\\
{n_{C{O_2}}} = \dfrac{{0,44}}{{44}} = 0,01\,mol\\
{V_{C{O_2}}} = 0,01 \times 22,4 = 0,224l\\
{n_{{H_2}}} = \dfrac{{0,04}}{2} = 0,02\,mol\\
{V_{{H_2}}} = 0,02 \times 22,4 = 0,448l\\
4)\\
a)\\
Fe + 2HCl \to FeC{l_2} + {H_2}\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{HCl}} = 0,2 \times 1 = 0,2\,mol\\
{n_{Fe}} = \dfrac{{{n_{HCl}}}}{2} \Rightarrow \text{ phản ứng vừa đủ } \\
{n_{FeC{l_2}}} = {n_{Fe}} = 0,1\,mol\\
{C_M}FeC{l_2} = \dfrac{{0,1}}{{0,2}} = 0,5M
\end{array}\)
\(\begin{array}{l}
{n_{{H_2}}} = {n_{Fe}} = 0,1\,mol\\
{V_{{H_2}}} = 0,1 \times 22,4 = 2,24l
\end{array}\)
