Mg + 2Hcl -> Mgcl2 + H2 (1)
Mgo + 2Hcl -> Mgcl2 + H2o (2)
nH2=V/22.4=3.36/22.4=0.15 mol
theo pt (1) nMg=nH2=0.15 mol
=>mMg=n.M=0.15.24=3.6g
=> %mMg=mMg/mhh.100%=3.6/11.6.100%=31.034%
=>%mMgo=100%-31.034%=68.966%
mH2=n.M=0.15.2=0.3 g
mdd=mhh+mHcl-mH2=11.6+653.7-0.3=665 g
nHcl=m/M=653.7/36.5=18 mol
theo 2 pt ta có
nMgcl2=1/4nHcl=1/4.18=4.5 mol
=mMgcl2=n.M=4.5.95 =427.5 g
=>%mdd=mMgcl2/mdd.100%=427.5/665.100%=64.3%