\(n_{KOH}=0,1.0,05=0,005\left(mol\right)\\ n_{HCl}=0,1.0,1=0,01\left(mol\right)\\ KOH+HCl\rightarrow KCl+H_2O\\ Vì:\dfrac{0,005}{1}< \dfrac{0,01}{1}\Rightarrow HCldư\\ a,n_{KOH}=n_{KCl}=0,005\left(mol\right)\\ n_{HCl\left(dư\right)}=0,01-0,005=0,005\left(mol\right)\\ \left[H^+\left(dư\right)\right]=\dfrac{0,005}{0,1+0,1}=0,025\left(M\right)\\ \left[K^+\right]=\dfrac{0,005}{0,1+0,1}=0,025\left(M\right)\\ \left[Cl^-\right]=0,025+0,025=0,05\left(M\right)\\ b,pH=-log\left[H^+\left(dư\right)\right]=-log\left[0,025\right]=1,602\)

\(a,m_{1đ.v.C}=\dfrac{1,9926.10^{-23}}{12}=0,16605.10^{-23}\left(g\right)\\ b,m_O=0,16605.10^{-23}.16=2,6568.10^{-23}\left(g\right)\\ m_{Fe}=56.0,16605.10^{-23}=9,2988.10^{-23}\left(g\right)\\ m_{Na}=23.0,16605.10^{-23}=3,81915.10^{-23}\left(g\right)\\ m_{Cl}=35,5.0,16605.10^{-23}=5,894775.10^{-23}\left(g\right)\\ m_{Zn}=65.0,16605.10^{-23}=10,79325.10^{-23}\left(g\right)\\ m_{Al}=27.0,16605.10^{-23}=4,48335.10^{-23}\left(g\right)\\ m_P=31.0,16605.10^{-23}=5,14755.10^{-23}\left(g\right)\)