nH2=0,15mol
PTHH: Fe+2HCL=>FeCl2+H2
0,15<-0,3<-0,15<-0,15
mFe tham gia phản ứng :
mFe=0,15.56=8,4g
CM (HCl)=n:V=0,3:0,05=6M
n(H2)=3,36/22,4=0,15
Fe + 2HCl--->FeCl2+H2
0,15....0,3.....................0,15
m(Fe) t/g p/u=0,15*56=8,4(g)
Cm(HCl)=0,3/(50/1000)=6 M