\(n_{H_2SO_4}=0,02.2=0,04\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,04----->0,04
=> \(m_{BaSO_4}=0,04.233=9,32\left(g\right)\)
H2SO4+Bacl2->BaSO4+2HCl
0,04---------------------0,04
n H2SO4=0,04 mol
=>m BaSO4=0,04.233=9,32g