$n_{CuO}=\dfrac{16}{80}=0,2mol \\m_{HNO_3}=6,3\%.142=8,946g \\⇒n_{HNO_3}=\dfrac{8,946}{63}=0,142mol \\PTHH :$
$CuO + 2HNO_3\to Cu(NO_3)_2+H_2O$
Theo pt : 1 mol 2 mol
Theo đbài : 0,2 mol 0,142 mol
Tỉ lệ : $\dfrac{0,2}{1}>\dfrac{0,142}{2}$
⇒Sau phản ứng CuO còn dư
Theo pt :
$n_{CuO\ pư}=n_{Cu(NO_3)_2}=\dfrac{1}{2}.n_{HNO_3}=\dfrac{1}{2}.0,142=0,071mol \\⇒m_{CuO\ pư}=0,071.80=5,68g \\m_{Cu(NO_3)_2}=0,071.188=13,348g \\m_{dd\ spu}=5,68+142=147,68g \\⇒C\%_{Cu(NO_3)_2}=\dfrac{13,348}{147,68}.100\%=9,04\%$
