Đáp án + Giải thích các bước giải:

$a\big)$

Đặt $n_{Na}=x(mol);n_{Ca}=y(mol)$

$\to 23x+40y=17,2(1)$

$2Na+2H_2O\to 2NaOH+H_2$

$Ca+2H_2O\to Ca(OH)_2+H_2$

Theo PT: $\frac{1}{2}x+y=n_{H_2}=\frac{8,96}{22,4}=0,4(2)$

Từ $(1)(2)\to x=0,4(mol);y=0,2(mol)$

$\to\%m_{Na}=\frac{0,4.23}{17,2}.100\%\approx 53,49\%$

$\to \%m_{Ca}=100\%-53,49\%=46,51\%$

$b\big)$

Theo PT: $n_{NaOH}=x=0,4(mol);n_{Ca(OH)_2}=y=0,2(mol)$

$\to\begin{cases} m_{NaOH}=0,4.40=16(g)\\ m_{Ca(OH)_2}=0,2.74=14,8(g) \end{cases}$

$\to m_{\rm dd\,kiềm}=16+14,8=30,8(g)$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
Ca + 2{H_2}O \to Ca{(OH)_2} + {H_2}\\
hh:Na(a\,mol),Ca(b\,mol)\\
{n_{{H_2}}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
\left\{ \begin{array}{l}
23a + 40b = 17,2\\
\dfrac{a}{2} + b = 0,4
\end{array} \right.\\
 \Rightarrow a = 0,4;b = 0,2\\
{m_{Na}} = 0,4 \times 23 = 9,2g\\
\% Na = \dfrac{{9,2}}{{17,2}} \times 100\%  = 53,5\% \\
\% Ca = 100 - 53,5 = 46,5\% \\
b)\\
{n_{NaOH}} = {n_{Na}} = 0,4mol\\
{m_{NaOH}} = 0,4 \times 40 = 16g\\
{n_{Ca{{(OH)}_2}}} = {n_{Ca}} = 0,2mol\\
{m_{Ca{{(OH)}_2}}} = 0,2 \times 74 = 14,8g\\
{m_{{\rm{dd}}}} = {m_{kl}} + {m_{{H_2}O}} - {m_{{H_2}}} = 17,2 + 120 - 0,4 \times 2 = 136,4g
\end{array}\)