a)
\(n_{NaOH}\)=0,2.5=1(mol)
\(n_{Ba\left(OH\right)_2}\)=0,2.2=0,4(mol)
\(NaOH+HCL\rightarrow NaCl+H_2O\)(1)
1 1 1 1 (mol)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\left(2\right)\)
0,4 0,8 0,4 0,8 (mol)
\(m_{NaCl}\)=1.58,5=58,5g
\(m_{BaCl_2}\)=0,4.208=83,2g
b)
\(m_{HCl}\)=36,5.(1+0,8)=65,7g
\(m_{ddHCl}\)=\(\frac{65,7.100}{12.4}\)=530g
\(V_{ddHCl}=\frac{530}{1,06}\)=500ml