Đáp án:

 \(\begin{array}{l}
3)\\
{C_{{M_{NaCl}}}} = \dfrac{1}{3}M\\
{C_{{M_{NaOH}}}} = \dfrac{1}{3}M\\
4)\\
{C_{{M_{CaS{O_4}}}}} = \dfrac{1}{3}M\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{1}{3}M
\end{array}\)

Giải thích các bước giải:

 \(\begin{array}{l}
3)\\
HCl + NaOH \to NaCl + {H_2}O\\
{n_{HCl}} = {C_M} \times V = 0,1 \times 1 = 0,1mol\\
{n_{NaOH}} = {C_M} \times V = 0,2 \times 1 = 0,2mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow NaOH\text{ dư}\\
{n_{NaO{H_d}}} = {n_{NaOH}} - {n_{HCl}} = 0,1mol\\
{n_{NaCl}} = {n_{HCl}} = 0,1mol\\
{C_{{M_{NaCl}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M\\
{C_{{M_{NaOH}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M\\
4)\\
{H_2}S{O_4} + Ca{(OH)_2} \to CaS{O_4} + 2{H_2}O\\
{n_{{H_2}S{O_4}}} = {C_M} \times V = 0,2 \times 1 = 0,2mol\\
{n_{Ca{{(OH)}_2}}} = {C_M} \times V = 0,1 \times 1 = 0,1mol\\
\dfrac{{0,1}}{1} < \dfrac{{0,2}}{1} \Rightarrow {H_2}S{O_4}\text{ dư}\\
{n_{{H_2}S{O_4}_d}} = {n_{{H_2}S{O_4}}} - {n_{Ca{{(OH)}_2}}} = 0,1mol\\
{n_{CaS{O_4}}} = {n_{Ca{{(OH)}_2}}} = 0,1mol\\
{C_{{M_{CaS{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M\\
{C_{{M_{{H_2}S{O_4}}}}} = \dfrac{n}{V} = \dfrac{{0,1}}{{0,1 + 0,2}} = \dfrac{1}{3}M
\end{array}\)

3.  PTHH: HCl+NaOH→NaCl+H2O

         nHCl=0,1.1=0,1(mol)

         nNaOH=0,2.1=0,2(mol)

ta có tỉ số: 0,1/1<0,2/1

⇒nHCl hết, nNaOH dư

theo PTHH: nNaCl=nHCl=0,1(mol)

Vdd sau phản ứng =100+200=300ml=3l

Cm nacl= 0,1/0,3=1/3(M)

theo PTHH: nNaOH pư=nHCl=0,1(mol)

             nNaOHdư=0,2-0,1=0,1(mol)

Cm naoh=0,1/0,3=1/3(M)

4. PTHH: H2SO4+Ca(OH)2→ CaSO4+2H2O

  nH2SO4=0,2.1=0,2(mol)

 nCa(OH)2=0,1.1=0,1(mol)

ta có tỉ số

0,2/1>0,1/1

⇒nH2SO4 dư ,nCa(OH)2 hết

theo PTHH:nCaSO4=nCa(OH)2=0,1(mol)

Vdd sau phản ứng=200+100=300ml=0,3l

 Cm CaSO4=0,1/0,3=1/3(M)

theo PTHH:nH2SO4 pư=nCa(OH)2=0,1(mol)

nH2SO4 dư =0,2-0,1=0,1(mol)

Cm H2SO4=0,1/0,3=1/3(M)

Chúc bạn học tốt