\(n_{NaOH}=0.4\cdot1.5=0.6\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_{2O}\)

\(0.6...........0.3......................0.3\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.3}{1.5}=0.2\left(l\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.3}{0.4+0.2}=0.5\left(M\right)\)

\(a)n_{H_2SO_4}=0,15.1=0,15mol\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,3                 0,15            0,15              0,15

\(C_{M\left(NaOH\right)}=\dfrac{0,3}{0,1}=3M\\ b)C_{M\left(Na_2SO_4\right)}=\dfrac{0,15}{0,1+0,15}=0,6M\)

\(a,H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=1.0,4=0,4\left(mol\right)\\ n_{NaOH}=0,4.2=0,8\left(mol\right)\\ b,V_{ddNaOH}=\dfrac{0,8}{0,5}=1,6\left(l\right)\\ c,n_{Na_2SO_4}=n_{H_2SO_4}=0,4\left(mol\right)\\ V_{ddNa_2SO_4}=0,4+1,6=2\left(l\right)\\ C_{MddNa_2SO_4}=\dfrac{0,4}{2}=0,2\left(M\right)\)

\(n_{NaOH}=0,02.2=0,04\left(mol\right)\\ 2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ a.n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,04}{2}=0,02\left(mol\right)\\ C_{MddH_2SO_4}=\dfrac{0,02}{0,08}=0,25\left(M\right)\\ b.\left[Na^+\right]=\dfrac{0,02.2}{0,02+0,08}=0,4\left(M\right)\\ \left[SO^{2-}_4\right]=\dfrac{0,02}{0,02+0,08}=0,2\left(M\right)\)

\(n_{H_2SO_4}=0,3.1=0,3\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

_______0,6<------0,3----------->0,3

=> V = \(\dfrac{0,6}{1}=0,6\left(l\right)\)

b) \(C_{M\left(Na_2SO_4\right)}=\dfrac{0,3}{0,6+0,3}=0,333M\)

\(n_{H_2SO_4}=1.0,3=0,3(mol)\\ 2NaOH+H_2SO_4\to Na_2SO_4+2H_2O\\ \Rightarrow n_{NaOH}=0,6(mol)\\ a,V_{dd_{NaOH}}=\dfrac{0,6}{1}=0,6(l)\\ b,n_{Na_2SO_4}=0,3(mol)\\ \Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,3}{0,6+0,3}=0,33M\)

1.n_CO2=0,1 (mol ) 

TH1: Số mol của CO₂ dư => Khối lượng muối khan tối đa tạo được là:

m_muối=0,1.84=8,4<9,5 (loại ) 

TH2: CO₂ hết 

Gọi số mol CO₂ tạo muối Na2CO3;NaHCO3 lần lượt là x, y

2NaOH+CO2→Na2CO3+H2O

NaOH+CO2→NaHCO3

Ta có : \(\left\{{}\begin{matrix}x+y=0,1\\106x+84y=9,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

⇒n_NaOH=2.0,05+0,05=0,15 (mol)

⇒CM_NaOH=\(\dfrac{0,15}{0,1}\)=1,5M

Câu 2 thật ra anh thấy chưa chặt chẽ nha!

2. Vì CO2 bị hấp thụ hoàn toàn

=> CO2 hết, NaOH dư

nCO2 = 0,1 (mol)

CO2 + 2NaOH -------> Na2CO3 + H2O

\(n_{Na_2CO_3}=n_{CO_2}=0,1\left(mol\right)\)

=> \(m_{Na_2CO_3}=0,1.106=10,6\left(g\right)\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2mol\\ a.MgCO_3+H_2SO_4->MgSO_4+H_2O+CO_2\\ 2NaOH+H_2SO_{\text{4 }}->Na_2SO_4+2H_2O\\ b.n_{H_2SO_4dư}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.80.0,1:40=0,1mol\\ n_{H_2SO_4\left(MgCO_3\right)}=0,2mol\\ c.C\%=\dfrac{98.0,3}{200}.100\%=14,7\%\\ V=0,2.22,4=4,48L\\ d.m_{ddsau}=200+16,8-44.0,2+80=288g\\ C\%_{Na_2SO_4}=\dfrac{40.0,1}{288}.100\%=1,39\%\\ C\%_{MgSO_4}=\dfrac{120.0,2}{288}.100\%=8,33\%\)

\(n_{MgCO_3}=\dfrac{16,8}{84}=0,2\left(mol\right)\)

\(n_{NaOH}=\dfrac{80}{40}=2\left(mol\right)\)

PTHH :

\(MgCO_3+H_2SO_4\rightarrow MgSO_4+H_2O+CO_2\uparrow\)

0,2              0,2                                         0,2 

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

2                 1                 1

Vậy có 0,2 mol H2SO4 phản ứng với MgCO3 

       có 1 mol H2SO4 phản ứng với NaOH 

\(m_{H_2SO_4}=1,2.98=117,6\left(g\right)\)

\(c,C\%_{H_2SO_4}=\dfrac{117,6}{200}.100\%=58,8\%\)

\(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(d,m_{Na_2SO_4}=1.142=142\left(g\right)\)

\(m_{ddNaOH}=\dfrac{80.100}{10}=800\left(g\right)\)

\(m_{ddH_2SO_4dư}=1.98:58,8\%\approx166,67\left(g\right)\)

\(m_{ddNa_2SO_4}=800+166,67=966,67\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{142}{966,67}.100\%\approx14,69\%\)