$m_{CuSO_4}=3,2\%.400=12,8g \\⇒n_{CuSO_4}=\dfrac{12,8}{160}=0,08mol \\n_{Ba}=\dfrac{27,4}{137}=0,2mol \\a.PTHH : \\Ba+H_2O\to Ba(OH)_2+H_2↑(1)$
$\text{Theo pt (1) :}$
$n_{H_2}=n_{Ba}=0,2mol$
$⇒V_{H_2}=0,2.22,4=4,48l$
$\text{b.Theo pt (1) :}$
$n_{Ba(OH)_2}=n_{Ba}=0,2mol$
$PTHH :$
$Ba(OH)_2 + CuSO_4\to BaSO_4↓+Cu(OH)_2↓(2)$
$\text{Theo pt : 1 mol 1 mol}$
$\text{Theo đbài : 0,2 mol 0,08 mol}$
Tỉ lệ : $\dfrac{0,2}{1}>\dfrac{0,08}{1}$
$\text{⇒Ba(OH)2 dư}$
-Chất rắn sau khi nung gồm BaSO4 và CuO
$\text{Theo pt (2) :}$
$n_{BaSO_4}=n_{Cu(OH)_2}=n_{CuSO_4}=0,08mol \\⇒m_{rắn}=n_{BaSO_4}+n_{Cu(OH)_2}=0,08.233+0,08.80=25,04g \\c.m_{dd\ spu}=27,4+400-0,2.2-0,08.233-0,08.98=400,52g \\n_{Ba(OH)_2\ pư}=n_{CuSO_4}=0,08mol \\⇒n_{Ba(OH)_2\ dư}=0,2-0,08=0,12g \\⇒m_{Ba(OH)_2\ dư}=0,12.171=20,52g \\⇒C\%_{Ba(OH)_2\ dư}=\dfrac{20,52}{400,52}.100\%=5,12\%$
