\(2Na\left(a\right)+2H_2O\rightarrow2NaOH\left(a\right)+H_2\)
Gọi nNa ban đầu là a (mol)
\(m_{ddNaOH10\%}=240.1,1=264\left(g\right)\)
\(\Rightarrow m_{NaOH10\%}=\dfrac{10.264}{100}=26,4\left(g\right)\)
\(\Rightarrow m_{NaOH}\) có trong dd mới (30%) = 26,4 + 40a (g)
\(m_{ddNaOH30\%}=23a+264\left(g\right)\) \(\Rightarrow m_{NaOH30\%}=\dfrac{\left(23a+264\right)30}{100}=6,9a+79,2\left(g\right)\)
Ta có pt: \(26,4+40a=6,9a+79,2\)
\(\Rightarrow a=1,59\)
\(\Rightarrow m_{Na}=1,59.23=36,579g\).