Xét 64g đồng.
Sau phản ứng, $m_{\text{tăng}}= 64.16,67\%=10,6688g=m_{O_2}$
$\Rightarrow n_{O_2}=\dfrac{10,6688}{32}=0,3344 mol$
$2Cu+O_2\to 2CuO$
$\Rightarrow n_{CuO}=2n_{O_2}=0,6688 mol$
$m_{CuO}=0,6688.80=53,344g$
$\Rightarrow m_{Cu\text{dư}}=64+10,6688-53,344=21,3248g$
$\%Cu=\dfrac{21,3248.100}{64+10,6688}=28,56\%$