a, \(n_{H^+}=n_{OH^-}=9.10^{-3}\left(mol\right)\Rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{\dfrac{9.10^{-3}}{2}}{0,05}=0,09M\)

b, \(\left[SO_4^{2-}\right]=\dfrac{4,5.10^{-3}}{0,05+0,15}=0,6M\)

\(\left[Na^+\right]=\dfrac{0,15.0,06}{0,05+0,15}=0,045M\)

\(\left[H^+\right]=\left[OH^-\right]=\dfrac{9.10^{-3}}{0,05+0,15}=0,045M\)

Đáp án D

0,05M

Đặt x, y là số mol HCl và  H 2 SO 4  trong 40 ml dung dịch A.

HCl + NaOH → NaCl +  H 2 O

H 2 SO 4 + 2NaOH →  Na 2 SO 4  + 2 H 2 O

Số mol NaOH: x + 2y = 1.60/1000 = 0,06 mol (1)

Khối lượng 2 muối : 58,5x + 142y = 3,76 (2)

Từ (1), (2), giải ra : x = 0,04 ; y = 0,01.

C M HCl  = 0,04/0,04 = 1(mol/l)

C M H 2 SO 4 = 0,01/0,04 = 0,25 (mol/l)

Gọi \(\left\{{}\begin{matrix}C_{M\left(A\right)}=aM\\C_{M\left(B\right)}=bM\end{matrix}\right.\)

Giả sử trộn 50ml dd A với 50ml dd B để thu được 100ml dd C

=> \(\left\{{}\begin{matrix}n_{NaOH}=0,05a\left(mol\right)\\n_{Ba\left(OH\right)_2}=0,05b\left(mol\right)\end{matrix}\right.\)

\(n_{BaSO_4}=\dfrac{9,32}{233}=0,04\left(mol\right)\)

n_H2SO4 = 0,035.2 = 0,07 (mol)

PTHH: Ba(OH)₂ + H₂SO₄ --> BaSO₄ + 2H₂O

              0,04<----0,04<-------0,04

            2NaOH + H₂SO₄ --> Na₂SO₄ + 2H₂O

              0,06<----0,03

=> \(\left\{{}\begin{matrix}0,05a=0,06\\0,05b=0,04\end{matrix}\right.\)

=> a = 1,2; b = 0,8 

20 ml dd A chứa n_NaOH = 0,02.1,2 = 0,024 (mol)

\(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)

PTHH: 2NaOH + Al₂O₃ --> 2NaAlO₂ + H₂O

             0,024-->0,012

            Ba(OH)₂ + Al₂O₃ --> Ba(AlO₂)₂ + H₂O

             0,188<---0,188

=> \(V_{dd.B}=\dfrac{0,188}{0,8}=0,235\left(l\right)=235\left(ml\right)\)

\(A.2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\\ n_{NaOH}=0,04.1=0,04mol\\ n_{H_2SO_4}=0,04:2=0,02mol\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,02}{0,05}=0,4M\)

Cho quỳ tím vào dd \(H_2SO_4\) rồi nhỏ từ từ dd \(NaOH\) vào. Đến khi thấy quỳ tím từ màu đỏ trở về màu tím thì đó là thời điểm \(H_2SO_4\) được trung hoà hoàn toàn.

nNa2O = 0,125 mol
a. Na2O + H2O --------> NaOH
0,125 mol ----------------> 0,125 mol
--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M
b. H2SO4 + 2NaOH ------> Na2SO4 + H2O
....0,0625 <---0,125 mol
--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g
--> mH2SO4(20%) = 6,125/20% = 30,625 g
suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

nNa2O = 0,125 mol

a. Na2O + H2O --------> NaOH

0,125 mol ----------------> 0,125 mol

--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M

b. H2SO4 + 2NaOH ------> Na2SO4 + H2O

....0,0625 <---0,125 mol

--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g

--> mH2SO4(20%) = 6,125/20% = 30,625 g

suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

nNa2O = 0,125 mol

a. Na2O + H2O --------> NaOH

0,125 mol ----------------> 0,125 mol

--> CM(NaOH) n/V = 0,125/ 0,25 = 0,5 M

b. H2SO4 + 2NaOH ------> Na2SO4 + H2O

....0,0625 <---0,125 mol

--> mH2SO4(nguyên chất) = 0,0625*98 = 6,125 g

--> mH2SO4(20%) = 6,125/20% = 30,625 g

suy ra V = m/D = 30,625 / 1,14 = 26,86 ml

Đáp án A

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O

0,05 →     0,025 ( mol)

n_CO2 =0,25 mol

H₂SO₄ + 2NaHCO₃ → Na₂SO₄ + 2H₂O + 2CO₂

0,125  ←                                                   0,25 (mol)

⇒ nH₂SO_{4 ban đầu} = 0,025 + 0,125 =0,15mol

⇒ C_{M H2SO4 ban đầu} = 1,5

Đáp án A.

Câu 7 : 

\(n_{H2SO4}=0,1.1=0,1\left(mol\right)\)

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

\(n_{NaOH}=2n_{H2SO4}=2.0,1=0,2\left(mol\right)\Rightarrow V_{ddNaOH}=\dfrac{0,2}{1}=0,2\left(l\right)=200\left(ml\right)\)

Câu 8 : 

\(n_{H2SO4}=0,5.0,7=0,35\left(mol\right)\)

Pt : \(H_2SO_4+2KOH\rightarrow K_2SO_4+H_2O\)

\(n_{KOH}=2n_{H2SO4}=2.0,35=0,7\left(mol\right)\)

\(\Rightarrow m_{ddKOH}=\dfrac{0,7.56}{12\%}.100\%=326,67\left(g\right)\)

\(\Rightarrow V_{ddKOH}=\dfrac{326,67}{1,15}=284,06\left(ml\right)\)

Câu 12 : 

a) \(2Cu+O_2\xrightarrow[]{t^o}2CuO\)

 \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2+2NaCl\)

\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(CuSO_4+Fe\rightarrow FeSO_4+Cu\downarrow\)

b) \(MgSO_4+2KOH\rightarrow Mg\left(OH\right)_2+K_2SO_4\)

\(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)

\(MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(MgCl_2+2AgNO_3\rightarrow Mg\left(NO_3\right)_2+2AgCl\)

\(Mg\left(NO_3\right)_2+Na_2CO_3\rightarrow MgCO_3+2NaNO_3\)

\(MgCO_3\xrightarrow[]{t^o}MgO+CO_2\)

c) \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(NaOH=HCl\rightarrow NaCl+H_2O\)

\(2NaCl+2H_2O\xrightarrow[cmn]{đpdd}2NaOH+H_2+Cl_2\)

\(Cl_2+H_2\xrightarrow[]{as}2HCl\)

\(HCl+Fe\rightarrow FeCl_2+H_2\)

\(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl\)

\(Fe\left(OH\right)_2+H_2SO_4\rightarrow FeSO_4+2H_2O\)

\(FeSO_4+BaCl_2\rightarrow FeCl_2+BaSO_4\)

\(FeCl_2+2AgNO_3\rightarrow Fe\left(NO_3\right)_2+2AgCl\)

\(Fe\left(NO_3\right)_2+Mg\rightarrow Mg\left(NO_3\right)_2+Fe\)

e) \(4Al+3O_2\xrightarrow[]{t^o}2Al_2O_3\)

\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

\(Al_2\left(SO_4\right)_3+6KOH\rightarrow2Al\left(OH\right)_3+3K_2SO_4\)

\(Al\left(OH\right)_3+3HNO_3\rightarrow Al\left(NO_3\right)_3+3H_2O\)

\(Al\left(NO_3\right)_2+Mg\rightarrow Mg\left(NO_3\right)_2+Âl\)

\(2Al+3Cl_2\xrightarrow[]{t^o}2AlCl_3\)

Bạn xem đề chỗ AlCl₃ ra Al₂(SO₄)₃ nhé

Câu 10 : 

\(n_{KOH}=\dfrac{5,6\%.400}{100\%.56}=0,4\left(mol\right)\)

\(n_{CuSO4}=\dfrac{16\%.300}{100\%.160}=0,3\left(mol\right)\)

Pt : \(2KOH+CuSO_4\rightarrow K_2SO_4+Cu\left(OH\right)_2\)

         0,4          0,3                                0,2

Xét tỉ lệ : \(\dfrac{0,4}{2}< \dfrac{0,3}{1}\Rightarrow CuSO_4dư\)

\(\Rightarrow m_{kt}=m_{Cu\left(OH\right)2}=0,2.98=19,6\left(g\right)\)

Câu 11 : 

\(n_{BaCl2}=0,1.1=0,1\left(mol\right)\)

Pt : \(BaCl_2+K_2CO_3\rightarrow BaCO_3+2KCl\)

\(n_{KCl}=2n_{BaCl2}=2.0,1=0,2\left(mol\right)\Rightarrow C_{MKCl}=\dfrac{0,2}{0,1+0,1}=1M\)

a) $n_{H_2SO_4}  = 0,35.2 = 0,7(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$

Theo PTHH : 

$n_{NaOH} = 2n_{H_2SO_4} = 1,4(mol) \Rightarrow C_{M_{NaOH}} = \dfrac{0,7}{0,2} = 3,5M$

b) $V_{dd\ sau\ pư}  = 0,2 + 0,35 = 0,55(lít)$
$C_{M_{Na_2SO_4}} = \dfrac{0,7}{0,55} = 1,27M$