`#AkaShi`
PTHH:
`BaCl_2 + H_2SO_4 -> BaSO_4↓ + 2HCl`
`\text{___________________________________}`
`->m_{BaCl_2}=m_{dd}xxC%=200xx10,4%=20,8 (g)`
`->n_{BaCl_2}=(m_{BaCl_2})/(M_{BaCl_2})=(20,8)/208=0,1 (mol)`
`->n_{H_2SO_4}=n_{BaCl_2}=0,1 (mol)`
`->V_{dd\ H_2SO_4}=(m_{H_2SO_4})/(C_{M})=(0,1)/(0,1)=1=1000ml (l)`
`->m_{dd\ H_2SO_4}=1000xx1,14=1140 (g)`
`->n_{BaSO_4↓}=n_{BaCl_2}=0,1 (mol)`
`->m_{dd}=1140+200-(0,1xx233)=1316,7 (g)`
`->n_{HCl}=2xxn_{BaCl_2}=1xx0,1=0,2 (mol)`
`->C%_{HCl}=(0,2xx36,5)/(1316,7)xx100%=0,55%`
`\text{___________________________________}`
`**` Đáp án:
`C%_{HCl}=0,55%`