Ta có: \(n_{Cu}=\dfrac{23,04}{64}=0,36\left(mol\right)\)
a. PTHH: 2Cu + O2 ---to---> 2CuO
Theo PT: \(n_{CuO}=n_{Cu}=0,36\left(mol\right)\)
=> \(m_{CuO}=0,36.80=28,8\left(g\right)\)
b. Theo PT: \(n_{O_2}=\dfrac{1}{2}.n_{Cu}=\dfrac{1}{2}.0,36=0,18\left(mol\right)\)
Do lấy dư 10% so với khối lượng phản ứng nên:
\(n_{O_2\left(lấydư\right)}=0,18.\dfrac{10\%}{100\%}=0,018\left(mol\right)\)
=> \(V_{O_2}=0,018.22,4=0,4032\left(lít\right)\)