PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Ba+2H_2O\rightarrow Ba\left(OH\right)_2+H_2\)
Ta có: 23nNa + 137nBa = 18,3 (1)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Ba}=0,2\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Na}=0,2\left(mol\right)\\n_{Ba}=0,1\left(mol\right)\end{matrix}\right.\)
Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{Ba\left(OH\right)_2}=n_{Ba}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{M_{NaOH}}=\dfrac{0,2}{0,2}=1\left(M\right)\\C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,1}{0,2}=0,5\left(M\right)\end{matrix}\right.\)
