Đáp án+Giải thích các bước giải:

`n_(CuO)=12/(64+16)=0,15(mol)` 

`a)`

`PTHH:CuO+2CH_3COOH->(CH_3COO)_2Cu+H_2O`

`b)`

Theo `PTHH`

`=>n_(CH_3COOH)=n_(CuO)*2=0,15*2=0,3(mol)`

`=>m_(ct)(CH_3COOH)=0,3*60=18(g)`

`=>m_(dd)(CH_3COOH)=18/10 *100=180(g)`

$\color{#FF00CC}{Mt} \color{#CC00CC}{ha} \color{#9900CC} {nh} \color{#6600CC}{18} \color{#3300CC}{0} \color{#0000CC} {7}$