Đáp án + Giải thích các bước giải:

$a\big)$

$n_{NaOH}=\frac{150.40\%}{40}=1,5(mol)$

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

Theo PT: $n_{Na}=n_{NaOH}=1,5(mol);n_{H_2}=\frac{1}{2}n_{Na}=0,75(mol)$

$\to x=m_{Na}=1,5.23=34,5(g)$

$\to z=V_{H_2}=0,75.22,4=16,8(l)$

$\to y=m_{NaOH}+m_{H_2}-m_{Na}=117(g)$

$b\big)$

$V_{dd\,NaOH}=\frac{m}{D}=\frac{150}{1,25}=120(ml)=0,12(l)$

$\to C_{M\,NaOH}=\frac{1,5}{0,12}=12,5M$

Em tham khảo nha :

\(\begin{array}{l}
a)\\
2Na + 2{H_2}O \to 2NaOH + {H_2}\\
{m_{NaOH}} = \dfrac{{150 \times 40}}{{100}} = 0060g\\
{n_{NaOH}} = \dfrac{{60}}{{40}} = 1,5mol\\
{n_{Na}} = {n_{NaOH}} = 0,15mol\\
{m_{Na}} = 0,15 \times 23 = 34,5g\\
{m_{{H_2}O}} = 150 - 34,5 = 115,5g\\
{n_{{H_2}}} = \dfrac{{{n_{NaOH}}}}{2} = 0,75mol\\
{V_{{H_2}}} = 0,75 \times 22,4 = 16,8l\\
b)\\
{V_{NaOH}} = \dfrac{{150}}{{1,25}} = 120ml = 0,12l\\
{C_{{M_{NaOH}}}} = \dfrac{{1,5}}{{0,12}} = 12,5M
\end{array}\)