nNa = \(\dfrac{6,9}{23}\) =0,3 mol
2Na + 2H2O ->2 NaOH + H2
0,3mol ->0,3mol->0,15mol
=>mNaOH = 0,3 . 40 = 12g
=> mdd = 6,9 + 50 - 0,15.2 = 56,6 g
=> C% = \(\dfrac{12}{56,6}\).100% = 21,2%
1)
\(n_{Na}=\dfrac{m}{M}=\dfrac{6,9}{23}=0,3\left(mol\right)\)
\(n_{H_2O}=\dfrac{m}{M}=\dfrac{50}{18}=\dfrac{25}{9}\left(mol\right)\)
\(\dfrac{n_{Na}}{2}< \dfrac{n_{H_2O}}{2}\)
\(\dfrac{0,3}{2}< \dfrac{\dfrac{25}{9}}{2}\)
=> H2O dư
\(2Na+2H_2O\xrightarrow[]{}2NaOH+H_2\)
2 : 2 : 2 : 1
0,3 0,3 0,3 0,6 (mol)
\(m_{NaOH}=n.M=0,3.40=1,2\left(g\right)\\ \)
\(m_{dd_{NaOH}}=m_{NaOH}+m_{H_2O}=1,2+50=60,2\left(g\right)\)
\(C\%_{NaOH}=\dfrac{m_{NaOH}}{m_{dd_{NaOH}}}.100\%=\dfrac{1,2}{60,2}.100\%\approx2\%\)