\(\text{oxit axit: } SO_2, SO_3, CO, CO_2, P_2O_5\\ \text{oxit bazơ: } CaO, MgO, Na_2O, K_2O, BaO \)

\(n_{HCl}=n_{H2SO4}=0,5.0,2=0,1(mol)\\ m_{HCl}=0,1.36,5=3,65g\\ m_{H_2SO_4}=0,1.98=9,8g\\ n_H=0,1+0,1.2=0,3(mol)\\ →n_{H_2O}=\frac{0,3}{2}=0,15(mol)\\ m_{H_2O}=0,15.18=2,7g\\ BTKL:\\ m_{muối}=12+3,65+9,8-2,7=22,75g\\ →B\)

sửa trên nha

\(n_{HCl}=n_{H2SO4}=0,5.0,2=0,1(mol)\\ m_{HCl}=0,1.36,5=3,65g\\ m_{H_2SO_4}=0,1.98=9,8g\\ n_H=0,1+0,1.2=0,3(mol)\\ →n_{H_2O}=\frac{0,3}{2}=0,15(mol)\\ m_{H_2O}=0,15.18=2,7g\\ BTKL:\\ m_{muối}=12+3,65+9,8-2,7=22,75g\\ →B\)

\(n_{NaOH}=0,1.0,01=0,001(mol)\\ \Rightarrow n_{OH^{-}}=0,001(mol)\\ n_{HCl}=0,03.0,2=0,006(mol)\\ \Rightarrow n_{H^{+}}=0,006(mol)\\ H^{+}+OH^{-}\to H_2O\\ 0,001<0,006\\ OH^{-} hêt; H^{+} dư\\ n_{H^{+}}=0,006-0,001=0,005(mol)\\ [H^{+}]=\frac{0,005}{0,1+0,2}=\frac{1}{60}M\\ \to pH=-log(\frac{1}{60})=1,77 \)

\(X(2p, n)\Leftrightarrow \begin{cases} 2p+n=58\\ p+n=20+e \end{cases}\Leftrightarrow \begin{cases} 2p+n=58\\ p+n=20+p \end{cases}\Leftrightarrow \begin{cases} p=e=20\\ n=19\end{cases} \to Kali\)

\(n_{H_2SO_4}=0,1.1=0,1(mol)\\ \text{Bảo toàn H}\\ n_{H_2SO_4}=n_{H_2}=0,1(mol)\\ V_{H_2}=0,1.22,4=2,24(l)\\ \to D\)

Sử dụng H2SO4 vì nó có tính hấp thụ cao

\(n_{H_2}=\frac{5,6}{22,4}=0,25(mol)\\ m_{H_2}=0,25.2=0,5(g)\\ BT H:\\ n_{HCl}=2n_{H_2}=0,25.2=0,5(mol)\\ m_{HCl}=0,5.36,5=18,25(g)\\ BTKL:\\ m_{hh}+m_{HCl}=m_{muối}+m_{H_2}\\ 15+18,25=m_{muối}+0,5\\ \to m_{muối}=32,75(g)\\ \to D\)

\(n_{HCl}=n_{H2SO4}=0,5.0,2=0,1(mol)\\ m_{HCl}=0,1.36,5=3,65(g)\\ m_{H2SO4}=0,1.98=9,8(g)\\ BTKL:\\ m_{muối}=12+3,65+9,8=25,45g\\ \to A\)