Đổi 1,8 tấn = 1800 kg
\(m_{FeS_2}=1800.60\%=1080\left(kg\right)\\ \Rightarrow n_{FeS_2}=\dfrac{1080}{120}=9\left(kmol\right)\)
BTNT S: \(n_{H_2SO_4}=2n_{FeS_2}=18\left(kmol\right)\)
=> \(m_{H_2SO_4}=18.98=1764\left(kg\right)\Rightarrow m_{ddH_2SO_4\left(98\%\right)}=\dfrac{1764}{98\%}=1800\left(kg\right)\)