Đáp án:

\(\begin{array}{l}
a)\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = 98g\\
b)\\
{m_{Mg{{(OH)}_2}}} = 1,16g\\
{C_M}MgC{l_2} = 0,875M\\
{C_M}KCl = 0,125\,M
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
{n_{KOH}} = {C_M} \times V = 0,2 \times 2 = 0,4\,mol\\
2KOH + {H_2}S{O_4} \to {K_2}S{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = \dfrac{{0,4}}{2} = 0,2\,mol\\
{m_{{H_2}S{O_4}}} = n \times M = 0,2 \times 98 = 19,6g\\
{m_{{\rm{dd}}{H_2}S{O_4}}} = \dfrac{{{m_{ct}} \times 100}}{{{C_\% }}} = \dfrac{{19,6 \times 100}}{{20}} = 98g\\
b)\\
{n_{MgC{l_2}}} = {C_M} \times V = 0,3 \times 1 = 0,3\,mol\\
{n_{KOH}} = {C_M} \times V = 0,02 \times 2 = 0,04\,mol\\
MgC{l_2} + 2KOH \to Mg{(OH)_2} + 2KCl\\
\dfrac{{0,3}}{1} > \dfrac{{0,04}}{2} \Rightarrow \text{ $MgCl_2$ dư}\\
{n_{MgC{l_2}}}\text{ phản ứng} = {n_{Mg{{(OH)}_2}}} = \dfrac{{0,04}}{2} = 0,02\,mol\\
{n_{MgC{l_2}}}\text{ dư} = 0,3 - 0,02 = 0,28\,mol\\
{V_{{\rm{dd}}spu}} = 0,3 + 0,02 = 0,32l\\
{n_{KCl}} = {n_{KOH}} = 0,04\,mol\\
{m_{Mg{{(OH)}_2}}} = n \times M = 0,02 \times 58 = 1,16g\\
{C_M}MgC{l_2} = \dfrac{n}{V} = \dfrac{{0,28}}{{0,32}} = 0,875M\\
{C_M}KCl = \dfrac{n}{V} = \dfrac{{0,04}}{{0,32}} = 0,125\,M
\end{array}\)

`n_(KOH)=0,2.2=0,4(mol)`

`a,`

`2KOH+H_2SO_4->K_2SO_4+2H_2O`

`0,4`              `0,2`

`m_(H_2SO_4)=0,2.98=19,6(g)`

`m_(dd   H_2SO_4)=\frac{19,6}{20%}=98(g)`

`b,`

`n_(MgCl_2)=0,3.1=0,3(mol)`

`n_(KOH)=0,02.2=0,04(mol)`

`2KOH+MgCl_2->Mg(OH)_2+2KCl`

`0,04`           `0,02`           `0,02`         `0,04`

Ta có tỉ lệ

`n_(KOH)=\frac{0,04}{2}<n_(MgCl_2)=\frac{0,3}{1}`

`=>KOH`  hết

`m_(Mg(OH)_2)=0,02.58=1,16(g)`

`C_(MKCl)=\frac{0,04}{0,32}=0,125M`

`C_(MMgCl_2)=\frac{0,3-0,02}{0,32}=0,875M`