PTHH:

\(AgNO_3+HCl->AgCl+HNO_3\)

--> kết tủa là AgCl

\(n_{AgCl}=\dfrac{m}{M}=\dfrac{14,35}{108+35,5}=0,1\left(mol\right)\)

Dựa theo PTHH: 

--> \(n_{HCl}=n_{AgNO_3}=n_{HNO_3}=n_{AgCl}=0,1\left(mol\right)\)

\(V_{HCl}=\dfrac{n}{C_M}=\dfrac{0,1}{2}=0,05\left(lít\right)\)

\(V_{AgNO_3}=\dfrac{n}{C_M}=\dfrac{0,1}{1}=0,1\left(lít\right)\)

b) Dung dịch thu được sau phản ứng là \(HNO_3\)

\(V_{ddsau}=V_{HCl}+V_{AgNO_3}=0,05+0,1=0,15\left(lít\right)\)

\(C_{M\left(HNO_3\right)}=\dfrac{n}{V}=\dfrac{0,1}{0,15}=\dfrac{2}{3}M\)