PTHH: S + $O_{2}$ -to-> $SO_{2}$
a,$n_{S}$ = $\frac{1,6}{32}$ = 0.05 mol
$n_{O_2}$ = $\frac{2,24}{22,4}$=0,1 mol
=> S hết, $O_{2}$ dư
=> $m_{O_2}$ dư = (0,1-0,05).16.2=1,6g
b,Theo bài ra:$n_{SO_2}$= $n_{S}$ = 0,05 mol
$m_{SO_2}$ = 0,05.(32+16.2)=3,2g