a)`m_{H_{2}SO_{4}}=(m_{dd}.C%)/(100%)=(98.20%)/(100%)=19,6(g)`
`n_{H_{2}SO_{4}}=m/M=(19,6)/98=0,2(mol)`
`PTHH:H_{2}SO_{4}+BaCl_{2}->BaSO_{4}↓+2HCl`
`(mol)` 0,2 0,2 0,2 0,4
b) `m_{BaSO_{4}}=n.M=0,2.233=46,6(g)`
c) `m_{HCl}=n.M=0,4.36,5=14,6(g)`
`m_{ddHCl}=m_{ddH_{2}SO_{4}}+m_{ddBaCl_{2}}-m_{BaSO_{4}}=98+200-46,6=251,4(g)`
`C%_{HCl}=\frac{m_{ct}}{m_{dd}}.100%=(14,6)/(251,4).100%≈5,8%`