Đáp án + Giải thích các bước giải:

 $1$ tấn $= 1000kg$

$m_{FeS_2} = (100\% - 19\%) . 1000 = 810(kg)$

$\Rightarrow n_{FeS_2} = \dfrac{810}{120} = 6,75(kmol)$
PTHH:
$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2 \uparrow$
$2SO_2 + O_2 \overset{V_2O_5,t^o}{\rightleftharpoons} 2SO_3$

$SO_3 + H_2O \longrightarrow H_2SO_4$
Theo PTHH:
$n_{SO_2(lt)} = 2n_{FeS_2} = 2 . 6,75 = 13,5(kmol)$

$\Rightarrow n_{SO_2(tt)} = 13,5 . 90\% = 12,15(kmol)$
$n_{SO_3(lt)} = n_{SO_2(tt)} = 12,15(kmol)$
$\Rightarrow n_{SO_3(tt)} = 12,15 . 64\% = 7,776(kmol)$
$n_{H_2SO_4(lt)} = n_{SO_3(tt)} = 7,776(kmol)$

$\Rightarrow n_{H_2SO_4(tt)} = 7,776 . 80\% = 6,2208(kmol)$

$\Rightarrow m_{H_2SO_4} = 6,2208 . 98 = 609,6384(kg) = 609638,4(g)$

$\Rightarrow m_{ddH_2SO_4} = \dfrac{609638,4 . 100\%}{73,5\%} = 829440(g)$

$\Rightarrow V_{H_2SO_4} = \dfrac{m}{D} = \dfrac{829440}{1,8} = 460800(ml) = 460,8(l)$

\(\begin{array}{l}
\rm 1\,tấn=1000\,kg\\
4FeS_2+11O_2\xrightarrow{t^o}2Fe_2O_3+8SO_2\\
2SO_2+O_2\xrightarrow{t^o,V_2O_5}2SO_3\\
SO_3+H_2O\to H_2SO_4\\
n_{FeS_2}=\frac{1000.(1-19\%)}{120}=6,75(kmol)\\
Bao\,toan\,S:\,n_{H_2SO_4}=2n_{FeS_2}=13,5(kmol)\\
\to n_{H_2SO_4(tt)}=13,5.90\%.64\%.80\%=6,2208(kmol)\\
\to m_{dd\,H_2SO_4}=\frac{6,2208.98}{73,5\%}=829,44(kg)=829440(g)\\
\to V_{dd\,H_2SO_4}=\frac{829440}{1,8}=460800(ml)
\end{array}\)