5 tấn = 5000 kg
$m_{FeS_2} = 5000.60\% = 3000(kg)$
$n_{FeS_2} = \dfrac{3000}{120}(kmol)$
$n_{FeS_2\ pư} = \dfrac{3000}{120}.80\% = 20(kmol)$
$n_{H_2SO_4} = 2n_{FeS_2} = 40(kmol)$
$m_{H_2SO_4} = 40.98 = 3920(kg)$
$m_{dd\ H_2SO_4} = \dfrac{3920}{60\%} = 6533,33(kg)$