a, PTPƯ: SO₃ + H₂O ---> H₂SO₄

n_SO3=\(\dfrac{2,24}{22,4}=0,1mol\)

1 mol SO₃ ---> 0,1 mol H₂SO₄

nên 0,1 mol SO₃ ---> 0,1 mol H₂SO₄

C_{M H2SO4}=\(\dfrac{0,1}{0,5}\)=0,2 M

b, PTPƯ: Zn + H₂SO₄ ---> ZnSO₄ + H₂

1 mol H₂SO₄ ---> 1 mol Zn

nên 0,1 mol H₂SO₄ ---> 0,1 mol Zn

m_Zn=0,1.65=6,5 g

a) \(n_{CO_2}=0,1\left(mol\right);n_{NaOH}=0,15\left(mol\right)\\ Tacó:\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,15}{0,1}=1,5\\ \Rightarrow Xảyracácphảnứng:\\ NaOH+CO_2\rightarrow NaHCO_3\\ 2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\\ Đặt:\left\{{}\begin{matrix}n_{NaHCO_3}=x\left(mol\right)\\n_{Na_2CO_3}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}x+y=0,1\left(BTNT\left(C\right)\right)\\x+2y=0,15\left(BTNT\left(Na\right)\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\\ \Rightarrow CM_{NaHCO_3}=CM_{Na_2CO_3}=\dfrac{0,05}{0,1}=0,5M\)

b) \(NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaOH}=0,15\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,15.36,5}{25\%}=21,9\left(g\right)\)

a. Ta có: \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{NaOH}=1,5.\dfrac{100}{1000}=0,15\left(mol\right)\)

Ta có: \(T=\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,15}{0,1}=1,5\left(1< 1,5< 1\right)\)

Vậy ta có PTHH:

\(CO_2+2NaOH--->Na_2CO_3+H_2O\left(1\right)\)

\(CO_2+NaOH--->NaHCO_3\left(2\right)\)

Gọi x, y lần lượt là số mol của Na₂CO₃ và NaHCO₃.

Theo PT₍₁₎: \(n_{CO_2}=n_{Na_2CO_3}=x\left(mol\right)\)

Theo PT₍₁₎: \(n_{NaOH}=2.n_{Na_2CO_3}=2x\left(mol\right)\)

Theo PT₍₂₎: \(n_{CO_2}=n_{NaOH}=n_{NaHCO_3}=y\left(mol\right)\)

Vậy, ta có HPT:

\(\left\{{}\begin{matrix}x+y=0,1\\2x+y=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,05\\y=0,05\end{matrix}\right.\)

\(\Rightarrow n_{dd_{sau.PỨ}}=0,05+0,05=0,1\left(mol\right)\)

Ta có: \(V_{dd_{sau.PỨ}}=V_{dd_{NaOH}}=\dfrac{100}{1000}=0,1\left(lít\right)\)

\(\Rightarrow C_{M_{sau.PỨ}}=\dfrac{0,1}{0,1}=1M\)

b. \(PTHH:NaOH+HCl--->NaCl+H_2O\left(3\right)\)

Theo PT₍₃₎: \(n_{HCl}=n_{NaOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,15.36,5=5,475\left(g\right)\)

Ta có: \(C_{\%_{HCl}}=\dfrac{5,475}{m_{dd_{HCl}}}.100\%=25\%\)

\(\Leftrightarrow m_{dd_{HCl}}=21,9\left(g\right)\)

[K+]=[Cl-]=0,25M

[KOH dư]=0,25M

b)   2KCl   +    H₂SO₄    ----------->K₂SO₄   + 2HCl

  0,05(mol)---->0,025(mol)

=>v_H2SO4=\(\frac{0,025}{1}\)=0,025(lít)

c)pH=-log(0,25)=0,602

(câu c mình không chắc chắn lắm nha bạn!!!)

câu a) làm sao vậy mn chỉ mik vs

\(n_{KOH}=0.1\cdot1=0.1\left(mol\right)\)

\(n_{HCl}=0.1\cdot0.5=0.05\left(mol\right)\)

\(KOH+HCl\rightarrow KCl+H_2O\)

\(0.05.......0.05.......0.05\)

Dung dịch D : 0.05 (mol) KOH , 0.05 (mol) KCl

\(\left[K^+\right]=\dfrac{0.05+0.05}{0.1+0.1}=0.5\left(M\right)\)

\(\left[Cl^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(\left[OH^-\right]=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

\(0.05.........0.025\)

\(V_{dd_{H_2SO_4}}=\dfrac{0.025}{1}=0.025\left(l\right)\)

Đáp án A

Gọi nồng độ ban đầu của H₂SO₄ là xM

n_H+ = 0,5.x.2 = x mol

[H⁺] = x/3 = 10^-3 suy ra x = 3.10^-3M