a, Ta có: \(n_{Na}=\dfrac{2,3}{23}=0,1\left(mol\right)\)
\(m_{HCl}=60.7,3\%=4,38\left(g\right)\Rightarrow n_{HCl}=\dfrac{4,38}{36,5}=0,12\left(mol\right)\)
PT: \(2Na+2HCl\rightarrow2NaCl+H_2\)
Xét tỉ lệ: \(\dfrac{0,1}{2}< \dfrac{0,12}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{Na}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.24,79=1,2395\left(l\right)\)
b, \(n_{HCl\left(pư\right)}=n_{NaCl}=n_{Na}=0,1\left(mol\right)\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\)
Ta có: m dd sau pư = 2,3 + 60 - 0,05.2 = 62,2 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{NaCl}=\dfrac{0,1.58,5}{62,2}.100\%\approx9,41\%\\C\%_{HCl}=\dfrac{0,02.36,5}{62,2}.100\%\approx1,17\%\end{matrix}\right.\)
