a) PTHH: \(2Cu+O_2\underrightarrow{t^o}2CuO\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Cu}=\frac{3,6}{64}=0,05625\left(mol\right)\\n_{O_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỷ lệ: \(\frac{0,05625}{2}< \frac{0,2}{1}\) \(\Rightarrow\) Oxi còn dư, tính theo Cu
\(\Rightarrow n_{O_2\left(pư\right)}=0,028125mol\) \(\Rightarrow n_{O_2\left(dư\right)}=0,171875\left(mol\right)\)
\(\Rightarrow m_{O_2\left(dư\right)}=0,171875\cdot32=5,5\left(g\right)\)
b) Theo PTHH: \(n_{CuO}=n_{Cu}=0,05625mol\)
\(\Rightarrow m_{CuO}=0,05625\cdot80=4,5\left(g\right)\)
