$nZn=6,5/65=0,1mol$
$nO2=3,36/22,4=0,15mol$
$a/$
$pthh:$
$2Zn+ O2→2ZnO$
$theo$ $pt$ $2 mol$ $1 mol$
$theo$ $đbài :$ $0,1mol$ $ 0,15mol$
ta có tỷ lệ :
$\frac{0,1}{2}<\frac{0,15}{1}$
⇒Sau phản ứng O2 dư
$theo$ $pt$
$n_{O2 pư}=1/2.n_{Zn}=1/2.0,1=0,05mol$
$⇒n_{O2 dư}=0,15-0,05=0,1mol$
$⇒m_{O2 dư}=0,1.32=3,2g$
$b/$
$theo$ $pt$
$n_{ZnO}=n_{Zn}=0,1mol$
$⇒m_{ZnO}=0,1.81=8,1g$
$c/$
$ZnO+2HCl→ZnCl2+H2$
$theo$ $pt$
$n_{HCl}=2.n_{ZnO}=2.0,1=0,2mol$
$⇒m_{HCl}=0,2.36,5=7,3g$