Đáp án:

Giải thích các bước giải:

`a.`

`m_{AgNO_3}=100.17%=17(g)`

`n_{AgNO_3}=17/170=0,1(mol)`

`m_{HCl}=300.3,65%=10,95(g)`

`n_{HCl}=\frac{10,95}{36,5}=0,3(mol)`

`AgNO_3+HCl->AgCl↓+HNO_3`

`\frac{0,1}1<\frac{0,3}1=>HCl` dư

`n_{AgCl}=n_{AgNO_3}=0,1(mol)`

`m_{AgCl}=0,1.143,5=14,35(g)`

`b.`

`m_{ddspư}=100+300-14,35=385,65(g)`

`n_{HCldư}=0,3-0,1=0,2(mol)`

`n_{HNO_3}=n_{AgNO_3}=0,1(mol)`

`C%_{HCl}=\frac{0,2.36,5}{385,65}.100=1,893%`

`C%_{HNO_3}=\frac{0,1.63}{385,65}.100=1,634%`

mAgNO3=100.17/100=17g

nAgNO3=17/170=0,1mol

mHCl=300.3,65/100=10,95g

nHCl=10,95/36,5=0,3mol

AgNO3+HCl→AgCl+HNO3

nAgNO3/1<nHCl/1

→HCl dư,tính theo AgNO3

mAgCl=0,1.143,5=14,35g

mdd sau pứ=100+300-14,35=385,65g

mHCl dư=(0,3-0,1).36,5=7,3g

C%HCl dư=7,3/385,65.100%=1,89%

mHNO3=0,1.63=6,3g

C%HNO3=6,3/385,65.100%=1,63%