Ta có: $n_{NH_3}=\frac{15}{224} ; n_{CuO}=0,2mol.$$\Rightarrow CuO$ dư.
$3CuO +2NH_3 \Rightarrow 3Cu +N_2+3H_2O$
$\frac{45}{48}$<------$ \frac{15}{224}$
a/ Ta có : $n_{CuO pứ}=\frac{45}{448} \Rightarrow m_{CuO} = 8,04g$
b/ $CuO +2HCl\Rightarrow CuCl_2+H_2O$
$0,1->0,2$
Ta có: $n_{CuOdư}=0,2-\frac{45}{448}=0,1mol$
$\Rightarrow n_{HCl}=0,2mol \Rightarrow V=0,1l=100ml.$