a) \(m_{ddAl_2\left(SO_4\right)_3}=200\cdot1,2=240\left(g\right)\)

    \(\Rightarrow m_{Al_2\left(SO_4\right)_3}=\dfrac{240\cdot28,5}{100}=68,4\left(g\right)\)

    \(\Rightarrow n_{Al_2\left(SO_4\right)_3}=0,2mol\)

    \(\Rightarrow\left\{{}\begin{matrix}n_{Al^{3+}}=2n_{Al_2\left(SO_4\right)_3}=0,4mol\\n_{SO^{2-}_4}=0,6mol\end{matrix}\right.\)

b) \(n_{HCl}=0,1\cdot3=0,3mol\) \(\Rightarrow n_{Cl^-}=0,3mol\)

    \(n_{HNO_3}=0,1\cdot1=0,1mol\) \(\Rightarrow n_{NO^-_3}=0,1mol\)

    \(\Sigma n_{H^+}=n_{HCl}+n_{HNO_3}=0,3+0,1=0,4mol\)

Đáp án C

n_MgSO4= 0,1 mol; n_Al2(SO4)3= 0,1 mol; V_dd= 0,5 lít

C_MgSO4=0,2M; C_MAl2(SO4)3= 0,2M

MgSO₄ → Mg²⁺+ SO₄^2-

0,2M                       0,2M

Al₂(SO₄)₃→ 2Al³⁺+ 3SO₄^2-

0,2M                        0,6M

[SO₄^2-] = 0,8M

a, \(\left[Al^{3+}\right]=0,2.2=0,4\left(M\right)\)

\(\left[SO_4^{2-}\right]=0,2.3=0,6\left(M\right)\)

b, \(\left[Mg^{2+}\right]=0,15.1=0,15\left(M\right)\)

\(\left[Cl^-\right]0,15.2=0,3\left(M\right)\)

a)m Al2(SO4)3=6,84.\(\dfrac{200}{100}\)=13,68g

=>n Al2(SO4)3=0,04 mol

b)n HCl=3.0,2=0,6 mol

a. \(m_{Al_2\left(SO_4\right)_3}=200.6,84\%=13,68\left(g\right)\)

\(n_{Al_2\left(SO_4\right)_3}=\dfrac{13.68}{342}=0,04\left(mol\right)\)

b. \(n_{HNO_3}=0,2.3=0,6\left(mol\right)\)

2)

nKOH = 0.15*2=0.3 mol

nHCl = 0.25*3=0.75 mol

KOH + HCl --> KCl + H2O

Bđ: 0.3____0.45

Pư : 0.3____0.3____0.3

Kt: 0______0.15___0.3

DD sau phản ứng : 0.15 mol HCl dư , 0.3 mol KCl

CM H⁺= 0.15/0.25=0.6M

CM Cl^- = 0.15/0.25=0.6 M

CM K⁺= 0.3/(0.15+0.25)=0.75M

CM Cl^-= 0.3/(0.15+0.25)= 0.75M

Đáp án: B.

3 N a 2 C O 3  +  A l 2 ( S O ₄ ) 3  + 3 H 2 O → 2 A l ( O H ) 3  ↓ + 3 C O 2  + 3 N a 2 C O 4

3 mol           1 mol

x mol           0,025 x 0,02 mol

⇒ x = 0,0015 mol

a)

$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
$[Ba^{2+}] = C_{M_{Ba(OH)_2}} = 0,1M$
$[OH^-] = 0,1.2 = 0,2M$

b)

$Al_2(SO_4)_3 \to 2Al^{3+} + 3SO_4^{2-}$
$[Al^{3+}] = 0,1.2 = 0,2M$
$[SO_4^{2-}] = 0,1.3 = 0,3M$

Đáp án C

Đáp án C