a) $Ba(OH)_2 + 2HCl \to BaCl_2 + 2H_2O$
$n_{HCl} = 0,2.0,5 = 0,1(mol)$
$n_{BaCl_2} = \dfrac{1}{2}n_{HCl} = 0,05(mol)$
$m_{BaCl_2} = 0,05.208 = 10,4(gam)$

b) $n_{Ba(OH)_2} = n_{BaCl_2} = 0,05(mol)$

$\Rightarrow V = \dfrac{0,05}{2} = 0,025(lít) = 25(ml)$

Đổi 200ml = 0,2 lít

Ta có: \(n_{HCl}=0,5.0,2=0,1\left(mol\right)\)

a. PTHH: 2HCl + Ba(OH)₂ ---> BaCl₂ + 2H₂O

Theo PT: \(n_{BaCl_2}=\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{BaCl_2}=0,05.208=10,4\left(g\right)\)

b. Theo PT: \(n_{Ba\left(OH\right)_2}=n_{BaCl_2}=0,05\left(mol\right)\)

=> \(V_{dd_{Ba\left(OH\right)_2}}=\dfrac{0,05}{2}=0,025\left(lít\right)=25\left(ml\right)\)

\(a.n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ a.Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\\ 0,25.......0,25............0,25..........0,25\left(mol\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,25}{0,1}=2,5\left(M\right)\\ b.m_{\downarrow}=m_{CaCO_3}=100.0,25=25\left(g\right)\)

Áp dụng ĐLBTKL, ta có :

nNaOH = 4/40 = 0.1 (mol)

PTHH: NaOH + HCl -> NaCl + H2O

Từ PTHH: nNaCl = nHCl = nNaOH = 0.1 (mol)

a) mNaCl = 0.1*(23+35.5) = 5.85(g)

b) mHCl = 0.1*(1+35.5) = 3.65(g)

C%ddHCl = 3.65/100 * 100% = 3.65%

SO2+Ba(OH)2->BaSO3+H2O

0,2-----0,2---------0,2--------0,2

n SO2=4,48\22,4=0,2 mol

=>Cm=0,2\0,25=0,8M

=>m BaSO3=0,2.217=43,4g

Đáp án A

+   S ơ   đ ồ   p h ả n   ứ n g :         C l 2 O 2 ⏟ 8 , 654     ( g ) +     M g ,     A l           Z n ⏟ 0 , 396     m o l → t 0 o x i t m u ố i ⏟ 23 , 246     ( g ) → V   m l H C l   1 M M g 2 + ,     A l 3 + Z n 2 + ,     C l - ⏟ T → 0 , 572   m o l B a ( O H ) 2 M g ( O H ) 2 ↓ +     P h ả n   ứ n g   c ủ a   T   v ớ i     B a ( O H ) 2 :                   M g + 2 O H - → M g ( O H ) 2                   Z n 2 + + 4 O H - → Z n O 2 2 - + 2 H 2 O                 A l 3 + + 4 O H - → A l O 2 - + 2 H 2 O +     n M g = x n Z n = y n A l = z ⇒ n Y = x + y + z = 0 , 396 m Y = 24 x + 65 y + 27 z = 23 , 246 - 8 , 654 2 x + 4 y + 4 z = n O H - = 1 , 144 ⇒ x = 0 , 22 y = 0 , 12 z = 0 , 056 +     m X = 71 n C l 2 + 32 n O 2 = 8 , 654 B T E :     2 n C l 2 + 4 n O 2 = 0 , 22 . 2 + 0 , 12 . 2 + 0 , 056 . 3 ⇒ n C l 2 = 0 , 034 n O 2 = 0 , 195 +     n H + = 2 n O 2 - = 2 . 2 n O 2 = 0 , 78     m o l ⇒ V H C l     1 M = 0 , 78     l í t = 780     m l