$n_{S}=50/32=1,5625mol$
$n_{SO2}=96/64=1,5mol$
$S+ O_{2}→SO_{2}$
$theo$ $pt :$ $1 mol$ $1 mol$
$theo$ $đbài :$ $1,5625mol$ $ 1,5mol$
⇒Sau phản ứng S dư.
$theo$ $pt :$
$n_{S pư}=n_{SO2}=1,5mol$
$⇒n_{S dư}=1,5625-1,5=0,0625mol$
$⇒m_{S dư}=0,0625.32=2g$
$n_{O2}=n_{SO2}=1,5mol$
$⇒V_{O2}=1,5.22,4=33,6g$