Đáp án:

$m_1=13,1(g)$

$m_2=237,2(g)$

Giải thích các bước giải:

$n_{NaOH}=\frac{20}{40}=0,5(mol)$

$n_{H_2}=\frac{3,36}{22,4}=0,15(mol)$

$2Na+2H_2O\to 2NaOH+H_2(1)$

$Na_2O+H_2O\to 2NaOH(2)$

Theo PT: $n_{NaOH(1)}=n_{Na}=2n_{H_2}=0,3(mol)$

$\to n_{NaOH(2)}=0,5-0,3=0,2(mol)$

Theo PT: $n_{Na_2O}=\frac{1}{2}n_{NaOH(2)}=0,1(mol)$

$\to m_1=m_{Na}+m_{Na_2O}=0,3.23+0,1.62=13,1(g)$

Bảo toàn KL:

$m_1+m_2=m_{dd\,NaOH}+m_{H_2}$

$\to 13,1+m_2=250+0,15.2$

$\to m_2=237,2(g)$

Em tham khảo nha:

\(\begin{array}{l}
2Na + 2{H_2}O \to 2NaOH + {H_2}(1)\\
N{a_2}O + {H_2}O \to 2NaOH(2)\\
{n_{{H_2}}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
{n_{Na}} = {n_{NaOH(1)}} = 2{n_{{H_2}}} = 0,3\,mol\\
{n_{NaOH}} = \dfrac{{20}}{{40}} = 0,5\,mol\\
{n_{NaOH(2)}} = 0,5 - 0,3 = 0,2\,mol\\
{n_{N{a_2}O}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{m_1} = {m_{Na}} + {m_{N{a_2}O}} = 0,3 \times 23 + 0,1 \times 62 = 13,1g\\
{n_{{H_2}O}} = 0,3 + 0,1 = 0,4\,mol\\
{m_2} = {m_{{H_2}O}} = 0,4 \times 18 = 7,2g
\end{array}\)