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`

a) m AgNO3 = C%.mdd/100=10.510/100=51(g) -> n AgNO3 = m/M=51/170 = 0.3 (mol)

m HCl = C%.mdd/100=91,25.20/100=18,25(g) -> n HCl = m/M=18,25/36,5 = 0.5 (mol)

HCl + AgNO3 --> AgCl + HNO3

1          1                    1           1        (mol)

0,5      0,3                                          (mol)

0,5/1 >0,3/1 --> HCl dư

Vậy tính theo n AgNO3

n AgCl = 0,3 (mol) ---) m AgCl = n.M = 0,3. 143,5 =43,05(g)

b) m dd sau p/ứng = m dd ban đầu - m kết tủa = 510 + 91,25 - 43,05 = 558,2 (g)

m HNO3 = n.M = 0,3.63=18,9(g)

C% HNO3 = mct . 100/mdd = 18,9.100/558,25=3,38%

m HCl dư = n.M = 0,2.36,5 = 7,3 (g)

C% HCl = mct.100/mdd = 7,3.100/558,25 = 1.30%