V_H2SO4= 200 ml = 0,2 (l)

m_KOH = \(\dfrac{98.40}{100}\) = 39,2 g=> n_KOH = \(\dfrac{39,2}{56}=0,7\)mol

Na₂CO₃ + H₂SO₄ -> Na₂SO₄ + H₂O + CO₂\(\uparrow\)

...............0,35mol<---0,35mol<------------0,35 mol

CO₂ +2 KOH -> K₂CO₃ + H₂O

0,35<--0,7mol

a) m_Na2SO4 = 0,35 . 142= 49,7 g

=>m_K2CO3 = 57,6 - 49,7 = 7,9 g

b) CM_H2SO4 = \(\dfrac{0,35}{0,2}\) = 1,75 M

\(m_{KOH}=39,2g\)

\(\Rightarrow n_{KOH}=\dfrac{m}{M}=0,7\left(mol\right)\)

- Gọi số mol mỗi muối thu được là x và y mol ( x,y > 0 )

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+CO_2\)

\(2KOH+CO_2\rightarrow K_2CO_3+H_2O\)

...2x...........x...............x.....................

\(KOH+CO_2\rightarrow KHCO_3\)

..y.............y...............y................

Ta có : \(m_M=m_{K2CO3}+m_{KHCO3}=138x+100y=57,6\left(I\right)\)

Mà \(n_{KOH}=2x+y=0,7\left(II\right)\)

- Giai 1 và 2 ta được : \(\left\{{}\begin{matrix}x=0,2\\y=0,3\end{matrix}\right.\) mol.

\(\Rightarrow\left\{{}\begin{matrix}m_{K2CO3}=n.M=27,6\\m_{KHCO3}=n.M=30\end{matrix}\right.\) g

b, Theo PTHH : \(n_{H2SO4}=n_{CO2}=x+y=0,5mol\)

\(\Rightarrow C_M=\dfrac{n}{V}=2,5M\)

PTHH: \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

            \(CO_2+KOH\rightarrow KHCO_3\)

            \(CO_2+2KOH\rightarrow K_2CO_3+H_2O\)

Gọi \(\left\{{}\begin{matrix}n_{KHCO_3}=a\left(mol\right)\\n_{K_2CO_3}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow100a+138b=57,6\)  (1)

Ta có: \(n_{KOH}=\dfrac{98\cdot48\%}{56}=0,84\left(mol\right)=a+2b\)  (2)

Từ (1) và (2) \(\Rightarrow\) Nghiệm âm

*Bạn xem lại đề !!

\(2NaBr+Cl_2\rightarrow2NaCl+Br_2\left(1\right)\\ m_{giảm}=m_{Br_2}-m_{Cl_2}\\ \Leftrightarrow n_{NaCl\left(1\right)}=n_{NaBr\left(1\right)}=\dfrac{13,35}{160-71}=0,15\left(mol\right)\\ \Rightarrow\%m_{NaBr}=\dfrac{103.0,15}{42,6}.100\approx36,268\%\\ \Rightarrow\%m_{NaCl}\approx63,732\%\)

Bổ sung:

\(C\%_{ddNaBr\left(trongA\right)}=\dfrac{0,15.103}{200}.100=7,725\%\\ C\%_{ddNaCl\left(trongA\right)}=\dfrac{42,6-0,15.103}{200}.100=13,575\%\)

a) \(n_{SO_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

\(n_{NaOH}=0,5.2=1\left(mol\right)\)

PTHH: 2NaOH + SO₂ --> Na₂SO₃ + H₂O

               1------->0,5------>0,5

             Na₂SO₃ + SO₂ + H₂O --> 2NaHSO₃

                0,1<-----0,1--------------->0,2

=> Thu được muối Na₂SO₃, NaHSO₃

\(\left\{{}\begin{matrix}m_{Na_2SO_3}=0,4.126=50,4\left(g\right)\\n_{NaHSO_3}=0,2.104=20,8\left(g\right)\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}C_{M\left(Na_2SO_3\right)}=\dfrac{0,4}{0,5}=0,8M\\C_{M\left(NaHSO_3\right)}=\dfrac{0,2}{0,5}=0,4M\end{matrix}\right.\)