Ta có : (Mg, Cu, Zn, Fe) + O2 -----to-----> Oxit
Bảo toàn khối lượng : \(m_{KL}+m_{O_2}=m_{oxit}\)
=> \(m_{KL}=10-0,3.32=0,4\left(g\right)\)
\(n_{O_2}=\frac{6,72}{22,4}=0,3(mol)\\ m_{O_2}=0,3.32=9,6(g)\\ BTKL:\\ m_{hh}+m_{O_2}=m_{oxit}\\ m_{hh}+9,6=10 \Rightarrow m_{hh}=0,4(g)\)