\(NaOH+HCl-->NaCl+H2O\)
\(n_{NaOH}=0,2.2=0,4\left(mol\right)\)
\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
==>NaOH dư.
dd sau pư là NaOH dư và NaCl
\(V_{dd}=200+100=300ml=0,3l\)
\(n_{NaOH}=n_{HCl}=0,2\left(mol\right)\)
\(n_{NaOH}dư=0,4-0,2=0,2\left(mol\right)\)
\(C_{M\left(NaOH\right)dư}=\frac{0,2}{0,3}=\frac{2}{3}\left(M\right)\)
\(n_{NaCl}=n_{HCl}=0,2\left(mol\right)\)
\(C_{M\left(NaCl\right)}=\frac{0,2}{0,3}=\frac{2}{3}\left(M\right)\)