`#AkaShi`
PTHH:
`BaCl_2+H_2SO_4->BaSO_4↓+2HCl`
`a)`
`->m_{BaCl_2}=m_{dd}xxC%=250xx10%=25\ (g)`
`->n_{BaCl_2}=(m_{BaCl_2})/(M_{BaCl_2})=25/208\ (mol)`
`->m_{H_2SO_4}=40xx15%=6\ (g)`
`->n_{H_2SO_4}=6/98=3/49\ (mol)`
Xét hết dư:
`n_{BaCl_2}=20/208 > n_{H_2SO_4}=6/98\ (mol)`
`->H_2SO_4` hết và `BaCl_2` dư
`->n_{BaSO_4↓}=n_{H_2SO_4}=3/49\ (mol)`
`->m_{BaSO_4↓}=233xx3/49=699/49\ (g)`
`\text{_____________________________}`
`b)`
Chất có sau pư là:
`{(HCl),(BaCl_2):}`
`->n_{BaCl_2\ pư}=n_{H_2SO_4}=3/49\ (mol)`
`->n_{BaCl_2\ dư}=25/208-3/49=601/10192\ (mol)`
`->n_{HCl}=2n_{H_2SO_4}=2xx3/49=6/49\ (mol)`
`->m_{dd}=250+40-699/49=275,63\ (g)`
`->C%_{BaCl_2}=(601/10192xx208)/(275,63)xx100%=4,45%`
`->C%_{HCl}=(36,5xx6/49)/(275,63)xx100%=1,62%`
