nNa= 2,3/23=0,1 mol
nH2O=47,8/18=239/0- mol
2Na + 2H2O --> 2NaOH +H2
0,1 0,1 mol
ta thấy nNa/2<nH2O/2
=>Na hết , H2O dư
=> mNaOH = 0,1*40=4 g
mdd sau = 2,3 + 47,8 -0,05*2=50 g
C% NaOH= 4*100/50=8%
PTHH: \(Na+H_2O\rightarrow NaOH+\dfrac{1}{2}H_2\uparrow\)
Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,1\left(mol\right)\\n_{H_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,1\cdot40=4\left(g\right)\\m_{H_2}=0,05\cdot2=0,1\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Na}+m_{H_2O}-m_{H_2}=50\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\dfrac{4}{50}\cdot100\%=8\%\)