$m_{FeS_2}=\dfrac{10^6 \times 80}{100}=8 \times 10^5(g)$

$n_{FeS_2}=\dfrac{8 \times 10^5}{120}=\dfrac{10^5}{15}(mol)$

Bảo toàn nguyên tố $S:$

$n_{H_2SO_4}=2n_{FeS_2}=\dfrac{2 \times 10^5}{15} =\dfrac{2 \times 10^5}{15}(mol)$

Hiệu suất $90\%:$

$n_{H_2SO_4,thực}=\dfrac{2 \times 10^5}{15} \times \dfrac{90}{100}=0.12 \times 10^5(mol)$

$V=\dfrac{m_{H_2SO_4}}{D}=\dfrac{0.12 \times 10^5 \times 98}{1.82}≈646.2(L)$

`m_{FeS_2\ t t}=1.80\%.90\%=0,72(tấn)`

`FeS_2->2H_2SO_4`

`120->196(tấn)`

`0,72->x(tấn)`

`->x=m_{H_2SO_4\ t t}={0,72.196}/{120}=1,176(tấn)=1176(kg)`

`->m_{dd\ H_2SO_4}={1176}/{95\%}=1237,89(kg)`

`->V_{dd\ H_2SO_4}={1237,89}/{1,82}=680,16(L)`