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

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

\(n_{CuSO_4}=\dfrac{200.16\%}{160}=0,2\left(mol\right)\)

PTHH :

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

0,2              0,4                  0,2               0,2 

\(m_{NaOH}=0,4.40=16\left(g\right)\)

\(m_{ddNaOH}=\dfrac{16.100}{10}=160\left(g\right)\)

\(c,m_{Na_2SO_4}=0,2.142=28,4\left(g\right)\)

\(m_{ddNa_2SO_4}=200+160-\left(0,2.98\right)=340,4\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{28,4}{240,4}.100\%\approx8,34\%\)

\(d,PTHH:\)

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

0,2               0,2 

\(m_{CuO}=0,2.80=16\left(g\right)\)

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

b, \(m_{CuSO_4}=200.16\%=32\left(g\right)\Rightarrow n_{CuSO_4}=\dfrac{32}{160}=0,2\left(mol\right)\)

\(n_{NaOH}=2n_{CuSO_4}=0,4\left(mol\right)\Rightarrow m_{ddNaOH}=\dfrac{0,4.40}{10\%}=160\left(g\right)\)

c, \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{0,2.142}{200+160-0,2.98}.100\%\approx8,34\%\)

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

\(n_{CuO}=n_{Cu\left(OH\right)_2}=0,2\left(mol\right)\Rightarrow m_{CuO}=0,2.80=16\left(g\right)\)

\(n_{FeCl_2}=\dfrac{150\cdot12.7\%}{127}=0.15\left(mol\right)\)

\(n_{NaOH}=\dfrac{350\cdot4\%}{40}=0.35\left(mol\right)\)

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

\(0.15...........0.3................0.15............0.3\)

\(m_{Fe\left(OH\right)_3}=0.15\cdot90=13.5\left(g\right)\)

\(m_{dd}=150+350-13.5=486.5\left(g\right)\)

\(C\%_{NaCl}=\dfrac{0.3\cdot58.5}{486.5}\cdot100\%=3.61\%\)

\(C\%_{NaOH\left(dư\right)}=\dfrac{\left(0.35-0.3\right)\cdot40}{486.5}\cdot100\%=0.4\%\)

\(Fe\left(OH\right)_2\underrightarrow{^{^{t^0}}}FeO+H_2O\)

\(0.15..........0.15\)

\(m_{FeO}=0.15\cdot72=10.8\left(g\right)\)

\(4Fe\left(OH\right)_2+O_2\underrightarrow{^{^{t^0}}}2Fe_2O_3+4H_2O\)

\(0.15.........................0.075\)

\(m_{Fe_2O_3}=0.075\cdot160=12\left(g\right)\)

Đáp án A

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

PTHH: \(FeCl_2+2KOH\rightarrow2KCl+Fe\left(OH\right)_2\downarrow\)

a) Ta có: \(\left\{{}\begin{matrix}n_{FeCl_2}=\dfrac{25,4}{127}=0,2\left(mol\right)\\n_{KOH}=\dfrac{28}{56}=0,5\left(mol\right)\end{matrix}\right.\) 

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,5}{2}\) \(\Rightarrow\) FeCl₂ p/ứ hết, KOH dư

\(\Rightarrow n_{Fe\left(OH\right)_2}=0,2\left(mol\right)\) \(\Rightarrow m_{Fe\left(OH\right)_2}=0,2\cdot90=18\left(g\right)\)

b) 

+) Nung trong không khí

PTHH: \(4Fe\left(OH\right)_2+O_2\xrightarrow[]{t^o}2Fe_2O_3+4H_2O\)

Theo PTHH: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_2}=0,1\left(mol\right)\) \(\Rightarrow m_{Fe_2O_3}=0,1\cdot160=16\left(g\right)\)

+) Nung trong chân không

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

Theo PTHH: \(n_{FeO}=n_{Fe\left(OH\right)_2}=0,2\left(mol\right)\) \(\Rightarrow m_{FeO}=0,2\cdot72=14,4\left(g\right)\)