a) Fe +2HCl----->.FeCl2 +H2
Ta có
n\(_{H2}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
Theo pthh
n\(_{Fe}=n_{H2}=0,3\left(mol\right)\)
m=0,3.56=16,8(g)
b) m\(_{HCl}=0,6.35,5=21,3\left(g\right)\)
m\(_{dd}=\frac{21,3.100}{7,3}=292,78\left(g\right)\)