`n_{BaCl_2}={2,08}/208=0,01(mol)`

`n_{AgNO_3}={1,7}/170=0,01(mol)`

`PTHH:BaCl_2+2AgNO_3->Ba(NO_3)_2+2AgCl`

ta có: `{0,01}/1 >{0,01}/2=>BaCl_2` dư

`n_{Ba(NO_3)_2}=n_{BaCl_2 pư}=1/2 n_{AgNO_2}=0,005(mol)`

`n_{BaCl_2 dư}=0,01-0,005=0,005(mol)`

`a.m_{BaCl_2 dư}=0,005.208=1,04(g)`

`m_{Ba(NO_3)_2}=0,005.261=1,305(g)`

`b.V_{dd sau}=0,03+0,07=0,1(l)`

`C_M BaCl_2 dư={0,005}/{0,1}=0,05(M)`

`C_M Ba(NO_3)_2 ={0,005}/{0,1}=0,05(M)`

`BaCl_2+2AgNO_3->Ba(NO_3)_2+2AgCl`

`a)`

`n_[BaCl_2]=[2.08]/208=0.01` (mol)

`n_[AgNO_3]=[1.7]/170=0.01` (mol)

Tỉ lệ: `[0.01]/1>[0.01]/2 => AgNO_3` hết còn lại `BaCl_2` dư

`n_[Ba(NO_3)_2]=1/2 n_[AgNO_3]=1/2 * 0.01=0.005` (mol)

`n_[AgCl]=n_[AgNO_3]=0.01` (mol)

`-> m_[Ba(NO_3)_2]=0.005 * 261=1.305` (g)

`-> m_[AgCl]=0.01 * 143.5=1.435` (g)

`b)`

Các chất còn lại trong dung dịch là `Ba(NO_3)_2` và `BaCl_2` dư.

`m_[ddls]=30+70=100` (ml)`=0.1` (lit)

`n_[BaCl_2] (text[dư])=0.01 - 0.005=0.005` (mol)

`C_M BaCl_2 text[dư] =[0.005]/[0.1]=0.05` (M)

`C_M Ba(NO_3)_2=[0.005]/[0.1]=0.05` (M)

`#tbh2009`