\(HCl+AgNO3-->AgCl\downarrow+HNO3\)
\(V_{HCl}=120.1,1=132\left(ml\right)=0,132\left(l\right)\)
\(nHCl=0,5.0,132=0,066\left(mol\right)\)
\(n_{AgNO3}=0,2.0,1=0,02\left(mol\right)\)
=> HCl dư..> dd sau pư gồm HCl dư và HNO3
\(m_{ddAgNO3}=\frac{200}{1,25}=160\left(g\right)\)
\(n_{AgCl}=n_{AgNO3}=0,02\left(mol\right)\)
\(m_{AgCl}=0,02.143,5=2,87\left(g\right)\)
\(m_{dd}\) sau pư =\(m_{ddHCl}+m_{ddAgNO3}-m_{AgCl}\)
\(=120+160-2,87=277,13\left(g\right)\)
\(n_{HCl}=n_{AgNO3}=0,02\left(mol\right)\)
\(n_{HCl}dư=0,066-0,02=0,046\left(mol\right)\)
\(m_{HCl}dư=0,046.36,5=1,679\left(g\right)\)
\(C\%_{HCl}dư=\frac{1,679}{277,13}.100\%=0,6\%\)
\(n_{HNO3}=n_{AgNO3}=0,02\left(mol\right)\)
\(m_{HNO3}=0,02.63=0,0252\left(g\right)\)
\(C\%_{HNO3}=\frac{0,0252}{277,13}.100\%=9,09\%\)
