nFe=\(\dfrac{16,8}{56}\)=0,3 mol
PTHH
Fe+2HCl→FeCl2+H2
a/Theo pt :
nHCl=2.nFe=2.0,3=0,6mol
⇒mHCl=0,6.26,5=21,9g
⇒C%HCl=mctmdd.100%=\(\dfrac{21,9}{200}.100\)=10,95%
b/Theo pt :
nH2=nFe=0,3mol
⇒VH2=0,3.22,4=6,72l
c/nFeCl2=nFe=0,3mol
⇒mFeCl2=0,3.127=38,1g
mH2=0,3.2=0,6g
mddspư=16,8+200−0,6=216,2g
⇒C%FeCl2=\(\dfrac{38,1}{216,2}\).100%=17,62%