Bài 1:
SO3 + H2O → H2SO4
Gọi x là sô smol của SO3
\(\Rightarrow m_{SO_3}=80x\left(g\right)\)
Theo pT: \(n_{H_2SO_4}tt=n_{SO_3}=x\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}tt=98x\left(g\right)\)
\(m_{ddH_2SO_4.20\%}=100\times1,14=114\left(g\right)\)
\(\Rightarrow m_{H_2SO_4.20\%}=114\times20\%=22,8\left(g\right)\)
\(m_{H_2SO_4.25\%}=m_{H_2SO_4}tt+m_{H_2SO_4.20\%}=98x+22,8\left(g\right)\)
\(m_{ddH_2SO_4.25\%}=m_{SO_3}+m_{ddH_2SO_4.20\%}=80x+114\left(g\right)\)
\(C\%_{ddH_2SO_4.25\%}=\frac{98x+22,8}{80x+114}\times100\%=25\%\)
\(\Leftrightarrow\frac{98x+22,8}{80x+114}=0,25\)
\(\Rightarrow98x+22,8=20x+28,5\)
\(\Leftrightarrow78x=5,7\)
\(\Leftrightarrow x=\frac{5,7}{78}=\frac{19}{260}\left(mol\right)\)
Vậy \(n_{SO_3}=\frac{19}{260}\left(mol\right)\Rightarrow m_{SO_3}=\frac{19}{260}\times80=5,85\left(g\right)\)