Đáp án:

`a)``3Cu+8HNO_3->3Cu(NO_3)_2+2NO+4H_2O`
`CuO+2HNO_3->Cu(NO_3)_2+H_2O`

`b)Cu=96%`

`CuO=4%`

Giải thích các bước giải:

Đề 02:

`a)`

Pthh:`3Cu+8HNO_3->3Cu(NO_3)_2+2NO+4H_2O`

`CuO+2HNO_3->Cu(NO_3)_2+H_2O`

$n_{NO}$`=`$\dfrac{13,44}{22,4}$`=0,6` mol

`b)`

Theo pthh:$n_{Cu}$`=`$n_{NO}$`.3/2=0,6.3/2=0,9` mol

$m_{Cu}$`=n.M=0,9.64=57,6` g

$m_{CuO}$`=`$m_{hh}$`-`$m_{Cu}$`=60-57,6=2,4` g

`%`$m_{Cu}$`=`$\dfrac{57,6}{60}$`.100%=96%`

`%`$m_{CuO}$`=100-96=4%`

 

Đáp án:

$a)$ PTHH:

 $3Cu+8HNO_3→3Cu(NO_3)_2+2NO↑+4H_2O$

 $CuO+2HNO_3→Cu(NO_3)_2+H_2O$

$b)$ $\%m_{Cu}=96\%$

 $\%m_{CuO}=4\%$

Giải thích các bước giải:

 $a)$ PTHH:

 $3Cu+8HNO_3→3Cu(NO_3)_2+2NO↑+4H_2O$

 $CuO+2HNO_3→Cu(NO_3)_2+H_2O$

 $b)$ Ta có: $n_{NO}=\dfrac{13,44}{22,4}=0,6(mol)$

→ $n_{Cu}=\dfrac{3}{2}×0,6=0,9(mol)$

⇒ $\%m_{Cu}=\dfrac{0,9×64}{60}×100\%=96\%$

⇒ $\%m_{CuO}=100\%-96\%=4\%$