goi: x la khoi luong cua Na
y la khoi luong cua Na2O
\(\Rightarrow n_{Na}=\dfrac{x}{23}\left(mol\right)\) va \(n_{Na_2O}=\dfrac{y}{62}\left(mol\right)\)
\(2Na+2H_2O\rightarrow2NaOH+H_2\left(1\right)\)
DE: \(\dfrac{x}{23}\rightarrow\) \(\dfrac{x}{23}\rightarrow\) \(\dfrac{x}{46}\) (mol)
\(Na_2O+H_2O\rightarrow2NaOH\left(2\right)\)
DE: \(\dfrac{y}{62}\rightarrow\) \(\dfrac{2y}{62}\) (mol)
ta co: \(V_{H_2}=22,4.n=\dfrac{x}{46}.22,4=6,72\)
\(\Rightarrow x=\dfrac{6,72.46}{22,4}=13,8\left(g\right)\)
\(\Rightarrow y=26,2-13,8=12,4\left(g\right)\)
A, \(m_{NaOH}=m_{NaOH\left(1\right)}+m_{NaOH\left(2\right)}\)
\(=40.\left(\dfrac{13,8}{23}+\dfrac{12,4.2}{62}\right)=40\left(g\right)\)
B, \(\%m_{Na}=\dfrac{13,8}{26,2}.100\approx52,67\%\)
\(\%m_{Na_2O}=100-52,67=47,33\text{%}\)
