\(m_{dd\left(tang\right)}=m_{H2\left(mat.di\right)}=5.2-4,9=0,3\left(mol\right)\)
\(\Rightarrow n_{H2}=0,15\left(mol\right)\Rightarrow n_{HCl}=2.0,15=0,3\left(mol\right)\)
BTKL:
\(m_{hh}=m_{HCl}+m_{muoi}+m_{H2}\)
\(\Rightarrow m_{muoi}=5,2-0,5.36,5-0,3=15,85\left(g\right)\)