m(HCl)=36,5%*20g/100%=7,3g cantitatea de HCl care se contine si in solutia de 36,5% si in cea diluata cu 30g, jumate din solutia obtinuta, adica 25g solutie, ceea ce constituie 3,65g de HCl
HCl + AgNO₃ = AgCl + HNO₃
m(AgCl)=m(HCl)*Mr(AgCl)/Mr(HCl)=(3,65g*143,5g/mol)/36,5g/mol=14,35g
Răspuns: m(HCl)=7,3g; m(AgCl)=14,35g