Đáp án+Giải thích các bước giải:

`a)` `n_{HCl}=0,1.1=0,1(mol)`

`n_{NaOH}=1,5.0,2=0,3(mol)`

`NaOH+HCl->NaCl+H_2O`

  0,3   >   0,1                      ( mol )

`->` dd A có môi trường kiềm

`b)` Gọi V(lít) là thể tích dd cần lấy

`->n_{HCl}=2V(mol)`

`->n_{H_2SO_4}=1,5V(mol)`

`n_H=n_{HCl}=n_{H_2SO_4}=2V+2.1,5V=5V`

`n_{NaOH(dư)}=0,3-0,1=0,2(mol)`

`n_{OH}=n_{NaOH}=0,2(mol)`

`H+OH->H_2O`

`->5V=0,2`

`->V=0,04(l)=40(ml)`

\(\begin{array}{l}
a)\\
100ml=0,1l;200ml=0,2l\\
n_{HCl}=1.0,1=0,1(mol);n_{NaOH}=1,5.0,2=0,3(mol)\\
NaOH+HCl\to NaCl+H_2O\\
n_{HCl}<n_{NaOH}\to NaOH\,\rm dư\\
\to \rm A\,có\,MT\,bazơ\\
b)\\
Dat\,V_{dd\,B}=x(l)\\
\to n_{HCl}=2x(mol);n_{H_2SO_4}=1,5x(mol)\\
\to n_{H^+}=2x+1,5x.2=5x(mol)\\
n_{NaOH(du)}=0,3-0,1=0,2(mol)\\
\to n_{OH^-}=0,2(mol)\\
n_{OH^-}=n_{H^+}\\
\to 0,2=5x\\
\to x=0,04\\
\to V_{dd\,B}=0,04(l)=40(ml)
\end{array}\)