\(KOH+HCl\rightarrow KCl+H_2O\\ n_{KOH}=0,25\left(mol\right);n_{HCl}=0,2\left(mol\right)\\ LTL:\dfrac{0,25}{1}>\dfrac{0,2}{1}\\ \Rightarrow KOHdư\\ m_{rắn}=m_{KOH\left(dư\right)}+m_{KCl}=\left(0,25-0,2\right).56+0,2.74,5=17,7\left(g\right)\\ \Rightarrow ChọnD\)