a,
`n_{NO}=\frac{2,24}{22,4}=0,1\ (mol)`
Phương trình phản ứng:
`3Cu + 8HNO_3 \to 3Cu(NO_3)_2 + 2NO↑ + 4H_2O` (1)
`CuO + 2HNO_3 \to Cu(NO_3)_2 + H_2O` (2)
`n_{Cu}=\frac{3}{2}n_{NO}=0,15\ (mol)`
`⇒m_{Cu}=0,15×64=9,6\ (g)`
`⇒m_{CuO}=11,68-9,6=2,08\ (g)`
`⇒\%m_{CuO}=\frac{2,08}{11,68}×100\%=17,8\%`
b,
`\sum n_{Cu(NO_3)_2}=n_{Cu}+n_{CuO}=0,15+\frac{2,08}{80}=0,176\ (mol)`
`⇒m_{Cu(NO_3)_2}=0,176×188=33,088\ (g)`