2Na + 2H2O → 2NaOH + H2↑
\(n_{Na}=\frac{2,3}{23}=0,1\left(mol\right)\)
a) Theo pT: \(n_{NaOH}=n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,1\times40=4\left(g\right)\)
Theo PT: \(n_{H_2}=\frac{1}{2}n_{Na}=\frac{1}{2}\times0,1=0,05\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,05\times2=0,1\left(g\right)\)
Ta có: \(m_{dd}saupư=2,3+197,8-0,1=200\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{4}{200}\times100\%=2\%\)
b) \(V_{dd}saupư=\frac{200}{1,08}=185,19\left(ml\right)=0,18519\left(l\right)\)
\(\Rightarrow C_{M_{NaOH}}=\frac{0,1}{0,18519}=0,54\left(M\right)\)
2Na + 2H2O \(\rightarrow\) 2NaOH + H2
nNa = \(\frac{m}{M}\) = \(\frac{2,3}{23}\) = 0,1 mol
Theo PTHH, ta có: nNaOH = \(\frac{0,1.2}{2}\) = 0,1 mol
\(\Rightarrow\)mNaOH = n.M = 0,1 . 40 = 4g
n\(H_2\) = \(\frac{0,1.1}{2}\) = 0,05 mol
\(\Rightarrow\) m\(H_2\) = n.M = 0,05 . 2 = 0,1 g
mdd (sau pư) = 2,3 + 197,8 - 0,1 = 200 g
Vậy: C% NaOH = \(\frac{m_{CT}}{m_{dd}}.100\%\) = \(\frac{4}{200}.100\%\) = 2%
b) Vdd (sau pư) = \(\frac{m}{D}\) = \(\frac{200}{1,08}\) = 185,2 ml = 0,1852 l
CM = \(\frac{n}{V}\) = \(\frac{0,1}{0,1852}\) = 0,54 mol/l
