2Na + H2SO4 → Na2SO4 + H2 (1)
\(n_{Na}=\frac{11,96}{23}=0,52\left(mol\right)\)
\(m_{ddH_2SO_4}=197,98\times1,1=217,778\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}=217,778\times4,5\%=9,8\left(g\right)\)
\(\Rightarrow n_{H_2SO_4}=\frac{9,8}{98}=0,1\left(mol\right)\)
Theo PT1: \(n_{Na}=2n_{H_2SO_4}\)
Theo bài: \(n_{Na}=5,2n_{H_2SO_4}\)
Vì 5,2 > 2 ⇒ Na dư
2Na + 2H2O → 2NaOH + H2 (2)
Theo pT1: \(n_{H_2}=n_{H_2SO_4}=0,1=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,1\times2=0,2\left(g\right)\)
Theo PT1: \(n_{Na}pư=2n_{H_2SO_4}=2\times0,1=0,2\left(mol\right)\)
\(\Rightarrow n_{Na}dư=0,52-0,1=0,42\left(mol\right)\)
Theo Pt2: \(n_{H_2}=\frac{1}{2}n_{NaOH}=\frac{1}{2}\times0,42=0,21\left(mol\right)\)
\(\Rightarrow m_{H_2}=0,21\times2=0,42\left(g\right)\)
Ta có: \(m_{dd}saupứ=11,96+217,778-0,2-0,42=229,118\left(g\right)\)
Theo pT2: \(n_{NaOH}=n_{Na}=0,42\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,42\times40=16,8\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{16,8}{229,118}\times100\%=7,33\%\)
