`#AkaShi`
PTHH:
`2HCl + Ba(OH)_2->BaCl_2+2H_2O`
`->m_{HCl}=m_{dd}xxC%=200xx14,6%=29,2\ (g)`
`->n_{HCl}=(m_{HCl})/(M_{HCl})=(29,2)/(36,5)=0,8\ (mol)`
`->m_{Ba(OH)_2}=m_{dd}xxC%=200xx17,1%=34,2\ (g)`
`->n_{Ba(OH)_2}=(m_{Ba(OH)_2})/(M_{Ba(OH)_2})=(34,2)/(171)=0,2\ (mol)`
Xét hết dư:
`->n_{HCl}=(0,8)/2=0,4 > n_{Ba(OH)_2}=0,2\ (mol)`
`->HCl` dư và `Ba(OH)_2` hết
Sau phản ứng gồm dung dịch:
`{(BaCl_2),(HCl\ dư):}`
`->m_{dd}=m_{dd\ HCl}+m_{dd\ Ba(OH)_2}=200+200=400\ (g)`
`->n_{BaCl_2}=n_{Ba(OH)_2}=0,2\ (mol)`
`->n_{HCl\ dư}=2xxn_{Ba(OH)_2}=0,2xx2=0,4\ (mol)`
`->n_{HCl\ dư}=0,8-0,4=0,4\ (mol)`
`->C%_{HCl\ dư}=(0,4xx36,5)/400xx100%=3,65%`
`->C%_{BaCl_2}=(0,2xx208)/400xx100%=10,4%`
