Đáp án:
$14,04\%;6,49\%$
Giải thích các bước giải:
$CuO+2HCl \to CuCl_2+H_2O\\m_{HCl}=100.18,25\%=18,25(g)\\n_{HCl}=\dfrac{18,25}{36,5}=0,5(mol)\\n_{CuO}=\dfrac{4}{64+16}=0,05(mol)\\n_{CuO}=0,05(mol) <\dfrac{n_{HCl}}{2}\\ \to HCl\ du\\mdd_{spu}=4+100=104(g)\\n_{HCl\ pu}=2.0,05=0,1(mol)\\n_{HCl\ du}=0,5-0,1=0,4(mol)\\ \%m_{HCl\ du}=\dfrac{0,4.36,5}{104\%}=14,04\%\\ n_{CuCl_2}=n_{CuO}=0,05(mol)\\\%m_{CuCl_2}=\dfrac{0,05.(64+35,5.2)}{104\%}=6,49\%$