64/
$1 tấn=1000000g$
$m_{FeS_2}=1000000.75\%=750000g$
$⇒n_{FeS_2}=750000/120=6250mol$
$⇒n_S=2.n_{FeS_2}=2.6250=12500mol$
$⇒m_S=12500.32=400000g$
65/
$n_{SO_2}=9,8/22,4=0,4375mol$
$H=50\%⇒n_{SO_2tt}=0,4375.50\%=0,21875mol$
$2SO_2+O_2\overset{t^o}\to 2SO_3$
Theo pt :
$n_{SO_3}=n_{SO_2}=0,21875mol$
$⇒m_{SO_3}=0,21875.80=17,5g$