Gọi nH2=a mol nCl2=b mol

nhh khí=0,2 mol=a+b

H2 +Cl2=>2HCl

a mol b mol

0,8b<=0,8b=>1,6b mol

a-0,8b 0,2b mol

nHCl=1,6b=0,3(1,6b+a-0,8b+0,2b)

=>a=0,1625 mol b=0,0375 mol

b)nKOH=0,1 mol

HCl+KOH=>KCl+H2O

0,06 mol =>0,06 mol

3Cl2 +6KOH=>5KCl+KClO3+3H2O

0,05mol<= 0,1 mol

mdd sau=0,2.0,1625.71+0,06.36,5+40=44,4975gam

nKCl=0,06+1/12=43/300mol=>C