\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(2..........................2................1\)
\(m_A=m_{Na}+m_{H_2O}-m_{H_2}=46+200-2=244\left(g\right)\)
\(m_{NaOH}=2\cdot40=80\left(g\right)\)
\(C\%NaOH=\dfrac{80}{244}\cdot100\%=32.78\%\)
Đề cho thế này sao tính đc C% nhỉ