Đáp án:

Giải thích các bước giải:

$n_{AgCl}=\dfrac{14,35}{143,5}=0,1(mol)$

$HCl+AgNO_3\to HNO_3+AgCl\downarrow$

Theo PT: $n_{HCl}=n_{AgCl}=0,1(mol)$

$\to C\%_{HCl}=\dfrac{0,1.36,5}{100}.100\%=3,65\%$

**Lấy 10g dung dịch axit:

$n_{HCl}=\dfrac{10.3,65\%}{36,5}=0,01(mol)$

$HCl+NaOH\to NaCl+H_2O$

Theo PT: $n_{NaOH}=n_{HCl}=0,01(mol)$

$\to m_{dd_{NaOH}}=\dfrac{0,01.40}{20\%}=2(g)$

$\to V_{dd_{NaOH}}=\dfrac{2}{1,2}\approx 1,67(ml)$

$\#phnn$

$\\$

`HCl+AgNO_3->AgCl+HNO_3`

`   0,1  ←   0,1  ←  0,1  →0,1`

`n_(AgCl)=(14,35)/(143,5)=0,1(mol)`

`m_(dd_(sau_PƯ))=100-14,35=85,65(g)`

`C%_(HCl)=(m_(ct))/(m_(dd_(sau_PƯ)))=(0,1.36,5)/(85,65)≈0,04%`

`HCl+NaOH->NaCl+H_2O`

` 0,1  → 0,1  →0,1    → 0,1`

`m(dd_(NaOH))=(0,1.40.100%)/(C%)=20(g)`

`V_(dd_(NaOH))=(m_(dd_(NaOH)))/D=20/(1,2)≈16,7(ml)`