\(PTHH:4FeS_2\left(\dfrac{20000}{3}\right)+11O_2\underrightarrow{t^o}8SO_2\left(\dfrac{40000}{3}\right)+2Fe_2O_3\)
\(2SO_2\left(\dfrac{40000}{3}\right)+O_2\underrightarrow{t^o}2SO_3\left(\dfrac{40000}{3}\right)\)
\(SO_3\left(\dfrac{40000}{3}\right)+H_2O\rightarrow H_2SO_4\left(\dfrac{40000}{3}\right)\)
\(m_{FeS_2}=80\%=0,8\) tấn = 800000 (g)
\(\Rightarrow n_{FeS_2}=\dfrac{800000}{120}=\dfrac{20000}{3}\left(mol\right)\)
\(m_{H_2SO_4}\)(lý thuyết) = \(\dfrac{40000}{3}.98=\dfrac{3920000}{3}\left(g\right)\)
Vì \(H\%=95\%\Rightarrow m_{H_2SO_4}\)(trên thực tế) = \(\dfrac{3920000}{3}.95\%=\dfrac{3724000}{3}\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_460\%}=\dfrac{\dfrac{100.3724000}{3}}{60}=\dfrac{18620000}{9}\left(g\right)\approx2,07\) (tấn).
