Đáp án:
A
Giải thích các bước giải:
`\(\begin{array}{l}`
`2Na + 2{H_2}O \to 2NaOH + {H_2}\\`
`{n_{Na}} = \dfrac{{3,45}}{{23}} = 0,15mol\\`
`{n_{{H_2}}} = \dfrac{{{n_{Na}}}}{2} = 0,075mol\\`
`{n_{NaOH}} = {n_{Na}} = 0,15mol\\`
`{m_{NaOH}} = 0,15 \times 40 = 6g\\`
`C{\% _{NaOH}} = \dfrac{6}{{{m_{{\rm{dd}}spu}}}} \times 100\% = 4\% \\`
` \Rightarrow {m_{{\rm{dd}}spu}} = \frac{{6 \times 100}}{4} = 150g\\`
` \Rightarrow {m_{Na}} + {m_{{H_2}O}} - {m_{{H_2}}} = 150\\`
` \Rightarrow {m_{{H_2}O}} = 150 - 3,45 + 0,075 \times 2 = 146,7g`
