a) \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b) \(n_{HCl}=\dfrac{75.7,3\%}{36,5}=0,15\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,075<-0,15--->0,075->0,075
=> \(V_{H_2}=0,075.22,4=1,68\left(l\right)\)
c) \(m_{Fe}=0,075.56=4,2\left(g\right)\)
\(m_{FeCl_2}=0,075.127=9,525\left(g\right)\)