a)PTHH: \(Ba\left(OH\right)+Na_2CO_3\rightarrow2NaOH+BaCO_3\downarrow\)

             \(BaCO_3\underrightarrow{t^o}BaO+CO_2\uparrow\)

Ta có: \(n_{Ba\left(OH\right)_2}=0,4\cdot0,2=0,08\left(mol\right)\)

\(\Rightarrow n_{NaOH}=0,16mol\) \(\Rightarrow C_{M_{NaOH}}=\dfrac{0,16}{0,4}=0,4\left(M\right)\) (Coi V_dd thay đổi không đáng kể)

b) Theo PTHH: \(n_{BaCO_3}=n_{Ba\left(OH\right)_2}=n_{BaO}=0,08mol\) \(\Rightarrow m_{BaO}=0,08\cdot153=12,24\left(g\right)\)

a. Ba(OH)₂ +Na₂CO₃ ➝ BaCO₃ + 2NaOH

BaCO₃ ➝ BaO + CO₂

nBa(OH)₂ = 0,08 mol

=> nNaOH = 2nBa(OH)₂ = 0,16 mol

=> C_M = 0,4 M

b) Bảo toàn Ba: nBaO = nBa(OH)₂ = 0,08 mol

=> m = 12,24 g

a) Đổi 300 ml = 0,3 l

Cm=\(\dfrac{n}{v}\) \(\Rightarrow\) n_NaOH= Cm.V = 0,3.1 = 0,3 Mol

PTHH: 2NaOH + FeCl₂ \(\rightarrow\) 2NaCl + Fe(OH)₂

\(\Rightarrow\) n_Fe = 0,3 Mol

\(\Rightarrow\) V_Fe = \(\dfrac{n}{Cm}\) = \(\dfrac{0,3}{0,1}\) = 3 lít

b) ta có n_{\(_{Fe\left(OH\right)_2}\)}= 0,3 Mol

\(\Rightarrow\) m_{\(_{Fe\left(OH\right)_2}\)}= n.M = 0,3.90 = 27 g

c) mk nghĩ phải cho V_NaCl mới giải đc

V = 300 ml = 0,3 (l)

=>n_NaOH = 0,3.1 = 0,3 mol

2NaOH + FeCl₂ ->2 NaCl + Fe(OH)₂

0,3------>0,15---->0,3--------0,15

a) V_FeCl2 = \(\dfrac{0,15}{0,1}\)= 1,5 (l)

b) m_Fe(OH)2 = 0,15 . 90 = 13,5 g

c) Vdd = 1,5+ 0,3 = 1,8 (l)

C_M(NaCl) = \(\dfrac{0,3}{1,8}\) = 0,1666 M

ⁿNaOH =0,3

FeCl₂ + 2NaOH \(\rightarrow\) Fe(OH)₂ + 2NaCl

0,15 \(\leftarrow\) 0,3 \(\rightarrow0,15\rightarrow\)0,3

a,

^VFeCl₂ = 3(l)

b

^mFe(OH)₂ = 0,15 x 90 =13.5

c,

^CM_NaCl = 1M

\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)

Có lẽ đề cho dd HCl 1M bạn nhỉ?

a, Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl}=0,3.1=0,3\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\), ta được HCl dư.

Theo PT: \(n_{HCl\left(pư\right)}=2n_{MgO}=0,2\left(mol\right)\Rightarrow n_{HCl\left(dư\right)}=0,1\left(mol\right)\)

⇒ m_{HCl (dư)} = 0,1.36,5 = 3,65 (g)

Pư không tạo H₂ bạn nhé.

b, \(NaOH+HCl\rightarrow NaCl+H_2O\)

\(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2+2NaCl\)

Theo PT: \(n_{Mg\left(OH\right)_2}=n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)

\(\Rightarrow n_{Mg\left(OH\right)_2}=0,1.58=5,8\left(g\right)\)

\(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

a) Pt : \(FeCl_2+2NaOH\rightarrow Fe\left(OH\right)_2+2NaCl|\)

               1              2                  1                 2

             0,25          0,5               0,25 

b) \(n_{Fe\left(OH\right)2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

⇒ \(m_{Fe\left(OH\right)2}=0,25.90=22,5\left(g\right)\)

c) \(n_{FeCl2}=\dfrac{0,5.1}{2}=0,25\left(mol\right)\)

100ml = 0,1l

\(C_{M_{ddFeCl2}}=\dfrac{0,25}{0,1}=2,5\left(M\right)\)

d) Pt : \(Fe\left(OH\right)_2\underrightarrow{t^o}FeO+H_2O|\)

                1               1          1

               0,25          0,25

\(n_{FeO}=\dfrac{0,25.1}{1}=0,25\left(mol\right)\)

⇒ \(m_{FeO}=0,25.72=18\left(g\right)\)

 Chúc bạn học tốt

Lần sau bạn đăng tách từng bài ra nhé.

Bài 1:

Ta có: \(n_{NaOH}=\dfrac{100.12\%}{40}=0,3\left(mol\right)\)

PT: \(FeCl_2+2NaOH\rightarrow2NaCl+Fe\left(OH\right)_2\)

a, Theo PT: \(n_{FeCl_2}=n_{Fe\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,15\left(mol\right)\)

\(\Rightarrow m_{FeCl_2}=0,15.127=19,05\left(g\right)\)

\(\Rightarrow C\%_{FeCl_2}=\dfrac{19,05}{200}.100\%=9,525\%\)

b, Ta có: m _{dd sau pư} = 200 + 100 - 0,15.90 = 286,5 (g)

\(n_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{0,3.58,5}{286,5}.100\%\approx6,13\%\)

c, Phần này mình coi như nung trong điều kiện không có không khí nhé.

PT: \(Fe\left(OH\right)_2\xrightarrow[\left(kckk\right)]{t^o}FeO+H_2O\)

Theo PT: \(n_{FeO}=n_{Fe\left(OH\right)_2}=0,15\left(mol\right)\Rightarrow m_{FeO}=0,15.72=10,8\left(g\right)\)

Bài 4:

Ta có: \(n_{CuCl_2}=0,1.1,5=0,15\left(mol\right)\)

PT: \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

a, \(n_{Cu\left(OH\right)_2}=n_{CuCl_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,15.98=14,7\left(g\right)\)

b, \(n_{NaOH}=2n_{CuCl_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{NaOH}=\dfrac{0,3}{2}=0,15\left(l\right)\)

c, \(n_{NaCl}=2n_{CuCl_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,3}{0,15+0,1}=1,2\left(M\right)\)

Bài 2:

Ta có: \(n_{KOH}=0,1.0,5=0,05\left(mol\right)\)

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

Theo PT: \(n_{MgSO_4}=n_{K_2SO_4}=n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{KOH}=0,025\left(mol\right)\)

a, \(m_{Mg\left(OH\right)_2}=0,025.58=1,45\left(g\right)\)

b, \(C_{M_{MgSO_4}}=\dfrac{0,025}{0,1}=0,25\left(M\right)\)

c, \(C_{M_{K_2SO_4}}=\dfrac{0,025}{0,1+0,1}=0,125\left(M\right)\)

Bài 3: Bạn bổ sung đề bài này nhé.

PTHH: \(2NaOH+CuSO_4\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\)

a) Ta có: \(n_{CuSO_4}=\dfrac{320\cdot20\%}{160}=0,4\left(mol\right)=n_{Cu\left(OH\right)_2}\) \(\Rightarrow m_{Cu\left(OH\right)_2}=0,4\cdot98=39,2\left(g\right)\)

b) Theo PTHH: \(n_{NaOH}=2n_{CuSO_4}=0,8mol\) \(\Rightarrow m_{ddNaOH}=\dfrac{0,8\cdot40}{10\%}=320\left(g\right)\)

c) Theo PTHH: \(n_{Na_2SO_4}=n_{CuSO_4}=0,4mol\) \(\Rightarrow m_{Na_2SO_4}=0,4\cdot142=56,8\left(g\right)\)

Mặt khác: \(m_{dd}=m_{ddNaOH}+m_{ddCuSO_4}-m_{Cu\left(OH\right)_2}=600,8\left(g\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{56,8}{600,8}\cdot100\%\approx9,45\%\) 

\(n_{CuSO_4}=0.2\cdot0.5=0.1\left(mol\right)\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(0.1.............0.2.................0.1..........0.1\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.1}{0.3+0.2}=0.2\left(M\right)\)

\(Cu\left(OH\right)_2\underrightarrow{^{^{t^0}}}CuO+H_2O\)

\(0.1.............0.1\)

\(m_{CuO}=0.1\cdot80=8\left(g\right)\)