Giải thích các bước giải:
$a)\\n_{H_2}=\dfrac{1,68}{22,4}=0,075\ (mol)\\Fe+2HCl\to FeCl_2+H_2\\CuO+2HCl\to CuCl_2+H_2O\\n_{Fe}=n_{H_2}=0,075\ (mol)\\\to m_{Fe}=0,075×56=4,2\ (g)\\\to m_{CuO}=14,2-4,2=10\ (g)\\b)\\n_{CuO}=\dfrac{10}{80}=0,125\ (mol)\\n_{HCl}=2×(0,075+0,125)=0,4\ (mol)\\\to C_{M\ HCl}=\dfrac{0,4}{0,1}=4\ (M)$