a, PT: \(2Na+2H_2O\rightarrow2NaOH+H_2\)
\(Ca+2H_2O\rightarrow Ca\left(OH\right)_2+H_2\)
Gọi: \(\left\{{}\begin{matrix}n_{Na}=x\left(mol\right)\\n_{Ca}=y\left(mol\right)\end{matrix}\right.\)
⇒ 23x + 40y = 16,6 (1)
Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}+n_{Ca}=\dfrac{1}{2}x+y=0,4\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Na}=0,2.23=4,6\left(g\right)\\m_{Ca}=0,3.40=12\left(g\right)\end{matrix}\right.\)
b, Theo PT: \(\left\{{}\begin{matrix}n_{NaOH}=n_{Na}=0,2\left(mol\right)\\n_{Ca\left(OH\right)_2}=n_{Ca}=0,3\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{NaOH}=0,2.40=8\left(g\right)\\m_{Ca\left(OH\right)_2}=0,3.74=22,2\left(g\right)\end{matrix}\right.\)
