nBaCl2=\(\frac{104.10\%}{208}=0,05mol\)
H2SO4 + BaCl2=> BaSO4 + 2HCl
0,05<------0,05----->0,05------>0,1
mdd H2SO4 =\(\frac{0,05.98.100}{7}=70\left(g\right)\)
m tủa = 0,05.233=11,65(g)
mdd = 70+104-11,65=162,35 (g)
C% HCl = \(\frac{0,1.36,5}{162,35}.100\%=2,248\%\)