Giải thích các bước giải:
`n_(AgNO_3)=0,02xx1=0,02(mol)`
`n_(HCl)=0,15xx0,5=0,075(mol)`
`AgNO_3+HCl->AgCl↓+HNO_3`
Có `(0,02)/1<(0,075)/1` nên `HCl` dư.
`n_(HNO_3)=n_(AgNO_3)=0,02(mol)`
`n_(HCl text( dư))=0,075-0,02=0,055(mol)`
`V_(text(dd sau p/ứ))≈20+150=170(ml)=0,17(l)`
`->C_(M(HNO_3))=(0,02)/(0,17)≈0,118(M)`
`C_(M(HCl text( dư)))=(0,055)/(0,17)≈0,3235(M)`
`m_(text(dd sau p/ứ))=20xx1,1+150xx1,05=179,5(g)`
`->C%_(HNO_3)=(0,02xx63)/(179,5)xx100%≈0,702%`
`C%_(HCl text( dư))=(0,055xx36,5)/(179,5)xx100%≈1,1184%`