$n H_2$=$\frac{1,344}{22,4}$=0,06 (mol)
$Cu+HCl→$ ko pứ
$Fe+2HCl→FeCl_2+H_2↑$
0,06 0,12 ←0,06 (mol)
%$m Fe$=$\frac{0,06.56}{13}$.100 % ≈25,85 %
%$m Cu$=100 % - 25,85 %=74,15 %
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$m ct HCl$=0,12.26,5=4,38 (g)
$mdd HCl$=$\frac{4,38.100}{15}$ =29,2 (g)
---------------------Nguyễn Hoạt------------------