Đáp án:

\(\begin{array}{l} a,\ V_{H_2}=2,24\ lít.\\ b,\ C\%_{NaOH}=7,66\%\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l} a,\ PTHH:2Na+2H_2O\to 2NaOH+H_2↑\\ n_{Na}=\dfrac{4,6}{23}=0,2\ mol.\\ Theo\ pt:\ n_{H_2}=\dfrac12n_{Na}=0,1\ mol.\\ ⇒V_{H_2}=0,1\times 22,4=2,24\ lít.\\ b,\ Theo\ pt:\ n_{NaOH}=n_{Na}=0,2\ mol.\\ m_{\text{dd spư}}=m_{Na}+m_{H_2O}-m_{H_2}=4,6+100-0,1\times 2=104,4\ g.\\ ⇒C\%_{NaOH}=\dfrac{0,2\times 40}{104,4}\times 100\%=7,66\%\end{array}\)

chúc bạn học tốt!

`n_(Na)=\frac{4,6}{23}=0,2(mol)`

`a,`

`2Na+2H_2O->2NaOH+H_2`

`0,2`                       `0,2`         `0,1`

`V_(H_2)=0,1.22,4=2,24(l)`

`b,`

`m_(H_2)=0,1.2=0,2(mol)`

`m_(NaOH)=0,2.40=8(g)`

`m_(dd)=4,6+100-0,2=104,4(g)`

`C%_(NaOH)=\frac{8}{104,4}.100=7,66%`