\(m_{FeS_2\left(lý.thuyết\right)}=\dfrac{44.80}{100}=35,2\left(tấn\right)\)
=> \(\)\(n_{FeS_2\left(lý.thuyết\right)}=\dfrac{35,2.10^6}{120}=\dfrac{88.10^4}{3}\left(mol\right)\)
Bảo toàn S: \(n_{H_2SO_4\left(lý.thuyết\right)}=\dfrac{176.10^4}{3}\left(mol\right)\)
=> \(n_{H_2SO_4\left(tt\right)}=\dfrac{176.10^4}{3}.70\%=\dfrac{1232.10^3}{3}\left(mol\right)\)
=> \(m_{H_2SO_4\left(tt\right)}=\dfrac{1232.10^3}{3}.98=\dfrac{120736.10^3}{3}\left(g\right)\)