Đáp án:
\(\begin{array}{l} C\%_{NaOH}=16,08\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} PTHH:2Na+2H_2O\to 2NaOH+H_2↑\\ n_{Na}=\dfrac{23}{23}=1\ mol.\\ Theo\ pt:\ n_{NaOH}=n_{Na}=1\ mol.\\ m_{\text{dd NaOH}}=500\times 1,2=600\ g.\\ \Rightarrow ∑m_{NaOH}=1\times 40+600\times 10\%=100\ g.\\ Theo\ pt:\ n_{H_2}=\dfrac{1}{2}n_{Na}=0,5\ mol.\\ \Rightarrow m_{\text{dd spư}}=m_{Na}+m_{\text{dd NaOH}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=23+600-0,5\times 2=622\ g.\\ \Rightarrow C\%_{NaOH}=\dfrac{100}{622}\times 100\%=16,08\%\end{array}\)
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