a) BaCl_2+H_2SO_4->BaSO_4+2HCl
b)
n_{BaCl_2}={320.10,4\%}/{208}=0,16(mol)
n_{BaSO_4}=n_{BaCl_2}=0,16(mol)
m_↓=m_{BaSO_4}=0,16.233=37,28(g)
c)
n_{HCl}=2n_{BaCl_2}=0,32(mol)
n_{H_2SO_4}=n_{BaCl_2}=0,16(mol)
m_{dd\ H_2SO_4}={0,16.98}/{19,6\%}=80(g)
m_{dd\ spu}=80+320-37,28=362,72(g)
->C\%_{HCl}={0,32.36,5}/{362,72}.100\%=3,22\%
