Đáp án:
Giải thích các bước giải:
`a.`
`m_{AgNO_3}=\frac{m_{dd}.C}{100}=\frac{510.10}{100}=51(g)`
`n_{AgNO_3}=m/M=51/170=0,3(mol)`
`m_{HCl}=\frac{m_{dd}.C}{100}=\frac{91,25.20}{100}=18,25(g)`
`n_{HCl}=m/M={18,25}/{36,5}=0,5(mol)`
`AgNO_3+HCl->AgCl↓+HNO_3`
`\frac{0,3}1<\frac{0,5}1=>HCl` dư
`n_{AgCl}=n_{AgNO_3}=0,3(mol)`
`m_{AgCl}=0,3.143,5=43,05(g)`
`b.`
`m_{ddspư}=m_{ddAgNO_3}+m_{ddHCl}-m_{↓}=510+91,25-43,05=558,2(g)`
`n_{HCldư}=n_{HCl}-n_{HClpư}=0,5-0,3=0,2(mol)`
`n_{HNO_3}=n_{AgNO_3}=0,3(mol)`
`C%_{HCl}=\frac{m_{ct}}{m_{dd}}.100=\frac{0,2.36,5}{558,2}.100=1,308%`
`C%_{HNO_3}=\frac{m_{ct}}{m_{dd}}.100=\frac{0,3.63}{558,2}.100=3,386%`
`#Devil`
