\(\text{nHCl = 0,15.1 = 0,15 mol}\)
\(\text{Fe + 2HCl → FeCl2 + H2}\)
.............0,15......................0,075 (mol)
\(\text{→ V H2 = 0,075.22,4 = 1,68 lít }\)
Fe+2HCl--->FeCl2+H2
n HCl=0,15.1=0,15(mol)
Theo pthh
n H2=1.2 n HCl=0,075(mol)
V H2=0,075.22,4=1,68(l)