Đáp án:
V=2,8l
C%=2%
Giải thích các bước giải:
\(\begin{array}{l}
Na + {H_2}O \to NaOH + \dfrac{1}{2}{H_2}\\
2Al + 6{H_2}O \to 2Al{(OH)_3} + 3{H_2}\\
{n_{Na}} = 0,1mol\\
{n_{Al}} = 0,05mol\\
\to {n_{{H_2}}} = \dfrac{1}{2}{n_{Na}} + \dfrac{3}{2}{n_{Al}} = 0,125mol\\
\to {V_{{H_2}}} = 0,125 \times 22,4 = 2,8l\\
{n_{NaOH}} = {n_{Na}} = 0,1mol\\
\to {m_{NaOH}} = 0,1 \times 40 = 4g\\
\to {m_{{\rm{ddX}}}} = {m_{Na}} + {m_{Al}} + {m_{{H_2}O}} - {m_{{H_2}}}\\
\to {m_{{\rm{ddX}}}} = 2,3 + 1,35 + 196,6 - 0,275 \times 2 = 199,7g\\
\to C{\% _{NaOH}} = \dfrac{4}{{199,7}} \times 100\% = 2\% \\
\end{array}\)
