a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15<--0,3<-----0,15<--0,15
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
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