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Đáp án:

\(\begin{array}{l}
1\,A\\
2\,B\\
3\,A\\
4\,B\\
5\,C\\
6\,C
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{n_{Na}} = \dfrac{{2,3}}{{23}} = 0,1\,mol\\
{n_{{H_2}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{V_{{H_2}}} = 0,05 \times 22,4 = 1,12l\\
2)\\
FeO + {H_2} \xrightarrow{t^0} Fe + {H_2}O\\
{n_{Fe}} = {n_{FeO}} = 0,1\,mol\\
{m_{Fe}} = 0,1 \times 56 = 5,6g\\
3)\\
F{e_2}{O_3} + 3{H_2} \xrightarrow{t^0} 2Fe + 3{H_2}O\\
{n_{Fe}} = \dfrac{{5,6}}{{56}} = 0,1\,mol\\
{n_{F{e_2}{O_3}}} = \dfrac{{0,1}}{2} = 0,05\,mol\\
{m_{F{e_2}{O_3}}} = 0,05 \times 160 = 8g\\
4)\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
{n_{Ca}} = \dfrac{4}{{40}} = 0,1\,mol\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,1\,mol\\
{m_{Ca{{(OH)}_2}}} = 0,1 \times 74 = 7,4g\\
5)\\
CuO + {H_2} \xrightarrow{t^0} Cu + {H_2}O\\
{n_{Cu}} = {n_{{H_2}}} = 0,15\,mol\\
{m_{Cu}} = 0,15 \times 64 = 9,6g\\
6)\\
2Al + 3{H_2}S{O_4} \to A{l_2}{(S{O_4})_3} + 3{H_2}\\
{n_{Al}} = \dfrac{{2,7}}{{27}} = 0,1\,mol\\
{n_{{H_2}}} = 0,1 \times \dfrac{3}{2} = 0,15\,mol\\
{V_{{H_2}}} = 0,15 \times 22,4 = 3,36l
\end{array}\)