$NH_3+2O_2→HNO_3+H_2O$
Ta có: $m_{ddHNO_3}=5(tấn)=5000(kg)$
$⇒m_{HNO_3}=5000.85/100=4250(kg)$
$⇒m_{HNO_3lt}=4250.100/75=≈5666,67(kg)$
$n_{HNO_3lt}≈5666,67/63≈89,947(kmol)$
$⇒n_{HNO_3lt}=n_{NH_3}≈89,947(kmol)$
$⇒V_{NH_3}≈89,947.22,4≈2014,814$( nghìn lít)