Oleum là $H_2SO_4·n_{SO_3}$; oleum chứa $12,5\text{%} SO3$ nên sẽ chứa $87,5\text{%}H_2SO_4$
$m_{H_2SO_4}=91⇒m_{H_2O}=9⇒n_{H_2O}=0,5$
$SO_3+H_2O→H_2SO_4$
$⇒m_{H_2SO_4}=0,5·98+91=140⇒m_{oleum}=\frac{100·140}{87,5}=160⇒m_{SO_3}=20⇒∑n_{SO_3}=0,5+\frac{20}{80}=0,75$
$FeS_2→2SO_3⇒n_{FeS_2}=0,375⇒m_{FeS_2}=0,375·(56+32,2)=45$