a) Ta có
nAgNO3=0.5 mol ; nHCl =0.6 mol
\(\Sigma\)vdd= 0.8 l
PTHH
AgNO3 + HCl--> AgCl +HNO3
0.5---------0.5-------0.5-------0.5 mol
=>nHCl dư= 0.1 mol=>CM(HCl) =0.125M
CM HNO3= 0.625M
md =500*1.2+300*1.5+0.5*143.5=978.25g
=> C% (HNO3)=3.22%
=>C%(HCl)=0.37 %
AgNO3 + HCl = AgCl + HNO3
0,5............0,5.......0,5.......0,5
nAgNO3=0,5 mol
nHCl =0,3.2=0,6 mol
Theo PTPU => HCl dư
nAgCl=0,5 mol
nHNO3=0,5 mol
=> Cm HNO3= 0,5:0,8=0,625M
C% HNO3= (0,5.63):(500.1,2+300.1,5-0,5.143,5)=3,22%