1.

Đưa giấy quỳ vào 2 dd. $HCl$ làm quỳ hoá đỏ, còn lại $BaCl_2$.

2.

a,

$MgSO_4+2NaOH\to Mg(OH)_2+Na_2SO_4$

b,

$n_{MgSO_4}=\dfrac{90.20\%}{120}=0,15(mol)$

$\Rightarrow n_{NaOH}=0,15.2=0,3(mol)$

$m_{dd NaOH}=0,3.40:10\%=120g$

c,

$n_{Mg(OH)_2}=n_{Na_2SO_4}=0,15(mol)$

$m_{dd\text{spứ}}=90+120-0,15.58=201,3g$

$\to C\%_{Na_2SO_4}=\dfrac{0,15.142.100}{201,3}=10,58\%$

3.

a,

$CuSO_4+Ba(OH)_2\to BaSO_4+Cu(OH)_2$

$Cu(OH)_2\buildrel{{t^o}}\over\to CuO+H_2O$

$n_{CuSO_4}=\dfrac{32}{160}=0,2(mol)$

$n_{Ba(OH)_2}=\dfrac{150.17,1\%}{171}=0,15(mol)$

$\Rightarrow n_{Cu(OH)_2}=n_{BaSO_4}=0,15(mol)$

$n_{CuO}=n_{Cu(OH)_2}=0,15(mol)$

$\to m_{\text{rắn}}=0,15.80+0,15.233=46,95g$

b,

$n_{CuSO_4\text{dư}}=0,2-0,15=0,05(mol)$

$m_{dd\text{spứ}}=32+150-0,15.98-0,15.233=132,35g$

$\to C\%_{CuSO_4}=\dfrac{0,05.160.100}{132,35}=6,04\%$