$PTHH_{}$: $2Cu_{}$+$O_{2}$ →$2CuO_{}$
(đk nhiệt độ nhé)
a) $n_{O2}$ =$\frac{V}{22,4}$ =$\frac{22,4}{22,4}$=$1_{}(mol)$ \
Theo PTHH, ta có: $n_{CuO}$ =$2n_{O2}$ =$2.1_{}$ =$2(mol)_{}$
⇒$m_{CuO}$ =$n_{CuO}$. $M_{CuO}$ =$2{}$ .${80}$=$160(g){}$
b)$V_{CuO}$ =$n_{CuO}$. $22,4{}$ =$2{}$ .$22,4{}$= $44,8(l){}$