Ta có: \(n_{NO}=0,1\left(mol\right)\)
Gọi: nCu = x (mol)
Theo ĐLBT e, có: 2x = 0,1.3 ⇒ x = 0,15 (mol)
\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)
Bạn tham khảo nhé!
\(n_{NO}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
Bảo toàn e :
\(n_{Cu}=\dfrac{3\cdot n_{NO}}{2}=\dfrac{3\cdot0.1}{2}=0.15\left(mol\right)\)
\(m=0.15\cdot64=9.6\left(g\right)\)