Đáp án:
\(\begin{array}{l}
{m_{N{a_2}O}} = 9,3g\\
{m_{{H_2}O}} = 2,7g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
4Na + {O_2} \to 2N{a_2}O\\
{n_{Na}} = \dfrac{{6,9}}{{23}} = 0,3mol\\
{n_{{O_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1mol\\
\dfrac{{0,3}}{4} < \dfrac{{0,1}}{1} \Rightarrow {O_2}\text{ dư}\\
{n_{N{a_2}O}} = \frac{{{n_{Na}}}}{2} = 0,15mol\\
N{a_2}O + {H_2}O \to 2NaOH\\
{m_{N{a_2}O}} = 0,15 \times 62 = 9,3g\\
{n_{{H_2}O}} = {n_{N{a_2}O}} = 0,15mol\\
{m_{{H_2}O}} = 0,15 \times 18 = 2,7g
\end{array}\)