a: \(HCl+NaOH\rightarrow NaCl+H_2O\)
\(n_{HCl}=0.3\cdot1=0.3\left(mol\right)\)
\(n_{NaOH}=0.2\cdot0.5=0.1\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Lập tỉ lệ : \(\dfrac{0.3}{1}>\dfrac{0.1}{1}\Rightarrow HCldư\)
\(m_{NaCl}=0.1\cdot58.5=5.85\left(g\right)\)