Na2O + H2O → 2NaOH (1)
a) \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)
Theo PT1: \(n_{NaOH}=2n_{Na_2O}=2\times0,2=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,4}{0,3}=1,33\left(M\right)\)
b) NaOH + HCl → NaCl + H2O (2)
\(m_{HCl}=36,5\times20\%=7,3\left(g\right)\)
\(\Rightarrow n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
Theo PT2: \(n_{NaOH}=n_{HCl}\)
Theo bài: \(n_{NaOH}=2n_{HCl}\)
Vì \(2>1\) ⇒ NaOH dư
Dung dịch thu được sau phản ứng gồm: NaOH dư và NaCl
\(m_{NaOH}=0,4\times40=16\left(g\right)\)
\(\Rightarrow m_{dd}saupư=16+36,5=52,5\left(g\right)\)
Theo PT2: \(n_{NaOH}pư=n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow n_{NaOH}dư=0,4-0,2=0,2\left(mol\right)\)
\(\Rightarrow m_{NaOH}dư=0,2\times40=8\left(g\right)\)
\(\Rightarrow C\%_{NaOH}dư=\dfrac{8}{52,5}\times100\%=15,24\%\)
Theo PT2: \(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{NaCl}=0,2\times58,5=11,7\left(g\right)\)
\(\Rightarrow C\%_{NaCl}=\dfrac{11,7}{52,5}\times100\%=22,29\%\)
