`a)`

`n_{HCl}={7,3}/{36,5}=0,2(mol);n_{Ba(OH)_2}=0,05.1=0,05(mol)`

`Ba(OH)_2+2HCl->BaCl_2+2H_2O`

`0,05->0,1->0,05(mol)`

`0,05<{0,2}/2->HCl` dư

`->m_{HCl\ du}=7,3-0,1.36,5=3,65(g)`

`b)`

`m_{BaCl_2}=0,05.208=10,4(g)`

`c)`

Dd sau chứa HCl dư và `BaCl_2` làm quỳ hóa đỏ

$a)\\ 2HCl+Ba(OH)_2\to BaCl_2+2H_2O\\ n_{HCl}=\dfrac{7,3}{36,5}=0,2(mol)\\ n_{Ba(OH)_2}=\dfrac{50}{1000}.1=0,05(mol)\\ \dfrac{n_{HCl}}{2}>n_{Ba(OH)_2}\to HCl\ du\ spu\\ n_{HCl(pu)}=2n_{Ba(OH)_2}=0,1(mol)\\ m_{HCl(du)}=(0,2-0,1).36,5=3,65(g)\\ b)\\ n_{BaCl_2}=n_{Ba(OH)_2}=0,05(mol)\\ m_{BaCl_2}=0,05.208=10,4(g)\\ c)\\ quy\ tim\ hoa\ do\ vi\ HCl\ du\ spu.$