n SO3=\(\dfrac{32}{80}\)=0,4 mol
n BaCl2=2.0,1=0,2 mol
BaCl2+SO3+H2O->BaSO4+2HCl
0,2---------0,2---------------0,2--------0,4
=>SO3 dư :
SO3+H2O->H2SO4
0,2--------------0,2
=>m BaSO4=0,2.233=46,6g
C% axit =\(\dfrac{0,4.35,6+0,2.98}{32+100.1,2-46,6}\).100=32,44%