$n_{Mg}=\frac{1,2}{24}=0,05mol;n_{O_2}=\frac{3,36}{22,4}=0,15mol$ $2Mg$ + $O_2$ $\rightarrow $ $2MgO$
$\frac{0,05}{2}$ $<$ $\frac{0,15}{1}$ Vậy $O_2$ dư
$0,05$ $\rightarrow $ $0,025$ $\rightarrow $ $0,05$
$n_{O_{2}dư}=0,15-0,025=0,125mol$
$m_{O_{2} dư}=0,125.32=4g$
Sản phẩm là $MgO$
$m_{MgO}=0,05.(24+16)=0,05.40=2g$