\(n_{NaOH}=0,015.2=0,03\left(mol\right)\Rightarrow n_{OH^-}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=0,01.1,5=0,015\left(mol\right)\Rightarrow n_{H^+}=0,03\left(mol\right)\)

PT ion : H⁺ + OH^- ----------> H₂O

Ta có n _H+ = n _OH- => dd sau phản ứng trung hòa nên có PH =7

Ta có:

\(n_{HCl\left(D\right)}=n_{NaOH}=\frac{0,1.15}{1000}=0,0015\left(mol\right)\)

\(PTHH:NaOH+HCl\rightarrow NaCl+H_2O\)

\(AgNO_3+HCl\rightarrow AgCl+HNO_3\)

\(\Rightarrow CM_D=\frac{0,0015}{0,01}=0,15M\)

\(n_{HCl\left(E\right)}=n_{AgCl}=\frac{2,87}{143,5}=0,02\left(mol\right)\)

\(\Rightarrow CM_E=\frac{0,02}{0,08}=0,25M\)

Giải hệ PT:

\(\left\{{}\begin{matrix}3a+b=10\\a+3b=0,6\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_A=0,3M\\CM_B=0,1M\end{matrix}\right.\)

Đáp án B

6NH₃+ Al₂(SO₄)₃+ 6H₂O→ 2Al(OH)₃+3 (NH₄)₂SO₄ (1)

NaOH + Al(OH)₃ → NaAlO₂+ H₂O (2)

Có  n_Al(OH)3= n_NaOH= 0,01.2= 0,02 mol

→ n_Al2(SO4)3= 1 2 . n_Al(OH)3= 0,01 mol

→ C_{M Al2(SO4)3}= 0,01/ 0,02= 0,5M

Có \(m_M=m_{Na}+m_{Cl}+m_{SO_4}=1,38+m_{Cl}+m_{SO_4}=3,76\)

\(\Rightarrow m_{Cl}+m_{SO_4}=2,38\)

\(\Rightarrow35,5n_{HCl}+96n_{H_2SO_4}=2,38\)

Lại có : \(n_O=n_{NaOH}=0,06\left(mol\right)\)

BtH : \(n_H=2n_{H_2O}=2n_O=n_{H\left(NaOH\right)}+n_{H\left(HCl\right)}+2n_{H\left(H_2SO_4\right)}\)

\(\Rightarrow n_{HCl}+2n_{H_2SO_4}=0,06\)

\(\Rightarrow n_{HCl}=0,04\left(mol\right)\)

\(\Rightarrow C_{MHCl}=1M\)

Có: \(n_{NaOH}=0,2.1=0,2\left(mol\right)\)

PT: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

______0,2_____0,1_______0,1 (mol)

a, \(V_{ddH_2SO_4}=\dfrac{0,1}{2}=0,05\left(l\right)\)

b, \(C_{M_{Na_2SO_4}}=\dfrac{0,1}{0,2+0,05}=0,4M\)

Bạn tham khảo nhé!

a, \(\left[Ca^{2+}\right]=\dfrac{0,15.0,5}{0,15+0,05}=0,375M\)

\(\left[Na^+\right]=\dfrac{0,05.2}{0,15+0,05}=0,5M\)

\(\left[Cl^-\right]=\dfrac{0,15.2.0,5+0,05.2}{0,15+0,05}=1,25M\)

b, \(\left[Fe^{3+}\right]=\dfrac{\dfrac{2.1,6}{400}}{1,5}\approx0,005M\)

\(\left[K^+\right]=\dfrac{\dfrac{2.6,96}{174}}{1,5}\approx0,053M\)

\(\left[SO_4^{2-}\right]=\dfrac{\dfrac{3.1,6}{400}+\dfrac{6,96}{174}}{1,5}\approx0,035M\)

nOH-=0,5.2.0,2=0,2 mol

mHCL=200.1,05=210g=> mHCL=210.10/100=21g

n H+=nHCL=21/36,5=0,57 mol =>nH+>nOH-=>dd thu được có tính axit

nBa(OH)2 = 0.2*0.5=0.1 mol

mdd_HCl= V*d=200*1.05=210g

=> mHCl =210*10/100=21g

=> nHCl =21/36.5=\(\dfrac{42}{73}\) \(\approx\) 0.575mol

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

\(\dfrac{42}{73}\) 0.1

Ta thấy HCl dư \(\dfrac{137}{730}\) mol

Vậy dd thu được vẫn còn axit.

a)

$BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$n_{BaCl_2} = 0,1 < n_{H_2SO_4} = 0,2$ nên $H_2SO_4$ dư

$n_{BaSO_4} = n_{BaCl_2} = 0,1(mol)$
$m_{BaSO_4} = 0,1.233 = 23,3(gam)$

b)

A gồm :

$HCl : 0,1.2 = 0,2(mol)$
$H_2SO_4\ dư : 0,2 - 0,1 = 0,1(mol)$
$V_{dd} = 0,1 + 0,1= 0,2(lít)$

$C_{M_{HCl}} = \dfrac{0,2}{0,2} = 1M$
$C_{M_{H_2SO_4}} = \dfrac{0,1}{0,2} = 0,5M$

c)

$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} = 2n_{H_2SO_4\ dư} = 0,2(mol)$
$m_{dd\ NaOH} = \dfrac{0,2.40}{15\%} = 53,33(gam)$