Dd mới có pH = 7 ⇒ Pư vừa đủ

Có: \(n_{Ba\left(OH\right)_2}=0,05\left(mol\right)\)

PT: \(Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4+2H_2O\)

______0,05_____0,05______0,05 (mol)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,05}{0,1}=0,5\left(l\right)\)

\(m_{BaSO_4}=0,05.233=11,65\left(g\right)\)

\(nhh=5,6/22,4=0,25mol\)

\(Gọi nO2=a, nO3=b\)

\(→nO=2a+3b\)

a+b=0,25 (1)

CO+O→CO2

H2+O→H2O

nhh = 13,44/22,4 = 0,6 mol

Vì nO = nCO + nH2 = 0,6

→ 2a + 3b = 0,6 (2)

Giải (1) và (2)

a = 0,15

b = 0,1

% nO3 = 0,1/0,25 . 100 % = 40 %

2C_nH_2n + ₁OH  + 2Na -> 2C_nH_{2n + 1}ONa + H₂

C_nH_{2n + 1}OH + 1,5nO₂ -> n_CO2 + (n + 1) H₂O

n_H2 = \(\dfrac{2,24}{22,4}=0,1(mol)=>nX=0,1.2=0,2(mol)\)

n_CO2 = \(\dfrac{4,48}{22,4}=0,2(mol)\)

\(n=\dfrac{0,2}{0,2}=1=>CTHH:CH4O\)

CTHH :CH3OH

Ta có: \(n_{H_2SO_4}=0,1.0,5=0,05\left(mol\right)\)

\(n_{HNO_3}=0,1.0,1=0,01\left(mol\right)\)

\(\Rightarrow n_{H^+}=0,05.2+0,01=0,11\left(mol\right)\)

Có: \(n_{OH^-}=n_{NaOH}+n_{KOH}=V_{ddA}.1+V_{ddA}.0,5=1,5V_{ddA}\)

PT ion: \(H^++OH^-\rightarrow H_2O\)

\(\Rightarrow1,5V_{ddA}=0,11\) \(\Rightarrow V_{ddA}=\dfrac{11}{150}\left(l\right)\)

Ta có: \(m_m=m_{Na^+}+m_{K^+}+m_{SO_4^{2-}}+m_{NO_3^-}=\dfrac{11}{150}.23+\dfrac{11}{150}.0,5.39+0,05.96+0,01.62\approx8,54\left(g\right)\)

Ta có: \(n_{NaOH}=0,15.0,02=0,003\left(mol\right)\)

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

___0,0015_____0,003 (mol)

\(\Rightarrow V_{H_2SO_4}=\dfrac{0,0015}{0,03}=0,05\left(l\right)=50\left(ml\right)\)

\(n_{OH^-}=0,15.10^{-2}\)

\(n_{H^+}=0,35.\left(0,02+0,01.2\right)=0,014\)

\(n_{H^+d\text{ư}}=0,014-0,15.10^{-2}=0,0125\Rightarrow\left[H^+\right]d\text{ư}=\dfrac{0,0125}{0,15+0,35}=0,025\)

\(\Rightarrow pH\approx1,6\)