a)
\(n_{BaCO_3}=\dfrac{3,94}{197}=0,02\left(mol\right)\)
PTHH: \(BaCO_3+2HCl\rightarrow BaCl_2+CO_2+H_2O\)
0,02------>0,04----->0,02---->0,02
\(m_{HCl}=0,04.36,5=1,46\left(g\right)\\ m_{dd.HCl}=\dfrac{1,46}{14,6\%}=10\left(g\right)\)
b)
\(m_{dd.sau.pư}=3,94+10-0,02.44=13,06\left(g\right)\\ C\%_{BaCl_2}=\dfrac{0,02.208}{13,06}.100\%=31,85\%\)