Em tham khảo nha :
Em xem lại đề câu 4 nha
\(\begin{array}{l}
6)\\
a)\\
N{a_2}O + {H_2}O \to 2NaOH\\
{m_{{H_2}O}} = 40 \times 1 = 40g\\
{n_{N{a_2}O}} = a\,mol\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 2a\,mol\\
C{\% _{NaOH}} = \dfrac{{{m_{NaOH}}}}{{{m_{N{a_2}O}} + {m_{{H_2}O}}}} \times 100\% = 15,27\% \\
\Rightarrow \dfrac{{40 \times 2a}}{{62a + 40}} = 0,1527\\
\Rightarrow a = 0,08mol\\
{m_{N{a_2}O}} = 0,08 \times 62 = 4,96g\\
b)\\
{n_{NaOH}} = 2{n_{N{a_2}O}} = 0,16mol\\
{C_{{M_{NaOH}}}} = \dfrac{{0,16}}{{0,04}} = 4M
\end{array}\)
