Em tham khảo nha:
\(\begin{array}{l}
5)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{{11,2}}{{56}} = 0,2\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_{F{e_2}{O_3}}} = 0,1 \times 160 = 16g\\
6)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = \dfrac{{9,2}}{{23}} = 0,4\,mol\\
{n_{{H_2}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{V_{{H_2}}} = 0,2 \times 22,4 = 4,48l\\
{n_{NaOH}} = {n_{Na}} = 0,4\,mol\\
{m_{NaOH}} = 0,4 \times 40 = 16g\\
7)\\
a)\\
{C_M}KCl = \dfrac{1}{{0,75}} = 1,33M\\
b)\\
{n_{CuS{O_4}}} = \dfrac{{400}}{{160}} = 2,5\,mol\\
{C_M}CuS{O_4} = \dfrac{{2,5}}{4} = 0,625M\\
c)\\
{C_M}MgC{l_2} = \dfrac{{0,5}}{{1,5}} = 0,33M\\
d)\\
{C_M}N{a_2}C{O_3} = \dfrac{{0,06}}{{1,5}} = 0,04M\\
8)\\
a)\\
{n_{KN{O_3}}} = 0,5 \times 2 = 1\,mol\\
{m_{KN{O_3}}} = 1 \times 101 = 101g\\
b)\\
{n_{CaC{l_2}}} = 0,25 \times 0,1 = 0,025\,mol\\
{m_{CaC{l_2}}} = 0,025 \times 111 = 2,775g
\end{array}\)
