a) BaO + H2O → Ba(OH)2
\(n_{BaO}=\frac{61,2}{153}=0,4\left(mol\right)\)
Theo pT: \(n_{Ba\left(OH\right)_2}=n_{BaO}=0,4\left(mol\right)\)
\(\Rightarrow C_{M_{Ba\left(OH\right)_2}}=\frac{0,4}{2,5}=0,16\left(M\right)\)
b) \(n_{NaOH}=\frac{4}{40}=0,1\left(mol\right)\)
\(n_{NaOH.0,2M}=3\times0,2=0,6\left(mol\right)\)
\(\Rightarrow n_{NaOH}mới=0,1+0,6=0,7\left(mol\right)\)
\(\Rightarrow C_{M_{NaOH}}mới=\frac{0,7}{3}=0,233\left(M\right)\)
c) 2Na + 2H2O → 2NaOH + H2
\(n_{Na}=\frac{2,3}{23}=0,1\left(mol\right)\)
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: \(m_{ddNaOH}=2,3+100-0,1=102,2\left(g\right)\)
\(\Rightarrow C\%_{NaOH}=\frac{4}{102,2}\times100\%=3,91\%\)
