`BaCl_2+H_2SO_4->BaSO_4+2HCl`
`n_{H_2SO_4}=(200.9,8%)/98=0,2mol`
`n_{BaCl_2}=(800.6,5%)/(137+35,5.2)=0,25mol`
`n_{H_2SO_4}<n_{BaCl_2}`
`n_{BaSO_4}=n_{H_2SO_4}=0,2mol`
`m_{BaSO_4}=0,2.(137+32+16.4)=46,6g`
`n_{BaCl_2du}=0,25-0,2=0,05mol`
`mdd_{spu}=200+800-46,6=953,4g`
`%m_{BaCl_2du}=(0,05.(137+35,5.2))/(953,4%)=1,09%`