Em tham khảo nha :
\(\begin{array}{l}
1)\\
a)\\
2Fe + 3C{l_2} \to 2FeC{l_3}\\
b)\\
{n_{Fe}} = \dfrac{m}{M} = \dfrac{{11,2}}{{56}} = 0,2mol\\
{n_{C{l_2}}} = \dfrac{3}{2}{n_{Fe}} = 0,3mol\\
{V_{C{l_2}}} = n \times 22,4 = 0,3 \times 22,4 = 6,72l\\
{n_{FeC{l_3}}} = {n_{Fe}} = 0,2mol\\
{m_{FeC{l_3}}} = n \times M = 0,2 \times 162,5 = 32,5g\\
2)\\
a)\\
2Zn + {O_2} \xrightarrow{t^0} 2ZnO\\
b)\\
{n_{{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
{n_{Zn}} = 2{n_{{O_2}}} = 0,2mol\\
{m_{Zn}} = n \times M = 0,2 \times 65 = 13g\\
{n_{ZnO}} = 2{n_{{O_2}}} = 0,2mol\\
{m_{ZnO}} = n \times M = 0,2 \times 81 = 16,2g\\
3)\\
a)\\
2Cu + {O_2} \xrightarrow{t^0} 2CuO\\
b)\\
{n_{CuO}} = \dfrac{m}{M} = \dfrac{8}{{80}} = 0,1mol\\
{n_{Cu}} = {n_{CuO}} = 0,1mol\\
{m_{Cu}} = n \times M = 0,1 \times 64 = 6,4g\\
{n_{{O_2}}} = \dfrac{{{n_{CuO}}}}{2} = 0,05mol\\
{V_{{O_2}}} = n \times 22,4 = 0,05 \times 22,4 = 1,12l
\end{array}\)