a) \(K_2O+H_2SO_4\rightarrow K_2SO_4+H_2O\)
b) \(m_{H_2SO_4}=\dfrac{147.20}{100}=29,4\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{29,4}{98}=0,3mol\)
\(\Rightarrow m_{K_2SO_4}=0,3.174=52,5g\)
c) \(m_{K_2O}=0,3.94=28,2g\)
d) \(m_{ddsaupư}=28,2+147=175,2g\)
\(\Rightarrow C\%dd_{K_2SO_4}=\dfrac{52,5}{175,2}.100\%\approx29,96\%\)