nMgO = 2/40=0.05 mol
mHCl = 50*20/100=10 g
nHCl = 10/36.5=20/73 mol
MgO + 2HCl --> MgCl2 + H2O
Bđ: 0.05___20/73
Pư : 0.05___0.1_____0.05
Kt: 0______127/730__0.05
mHCl dư = 127/730*36.5 = 6.35 g
mMgCl2 = 4.75 g
mdd sau phản ứng = 2 + 50 = 52 (g)
C%HCl dư = 6.35/52*100% = 12.21%
%MgCl2 = 4.75/52*100% = 9.13%