\(a,PTHH:FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3\downarrow+3NaCl\\ \Rightarrow n_{Fe\left(OH\right)_2}=n_{FeCl_3}=\dfrac{10,7}{107}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{FeCl_3}}=0,1\cdot162,5=16,25\left(g\right)\\ \Rightarrow m_{dd_{FeCl_3}}=\dfrac{16,25\cdot100\%}{5\%}=325\left(g\right)\\ b,n_{NaOH}=n_{NaCl}=3n_{Fe\left(OH\right)_3}=0,3\left(mol\right)\\ \Rightarrow m_{NaOH}=0,3\cdot40=12\left(g\right)\\ m_{NaCl}=0,3\cdot58,5=17,55\left(g\right)\\ \Rightarrow m_{dd_{NaCl}}=325+150-10,7=464,3\left(g\right)\\ \Rightarrow C\%_{dd_{NaCl}}=\dfrac{17,55}{464,3}\cdot100\%\approx3,78\%\)

a) n Fe = 28/56 = 0,5(mol)

$Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH :

n HCl = 2n Fe = 1(mol)

=> m dd HCl = 1.36,5/10% = 365(gam)

b)

n FeCl2 = n H2 = n Fe = 0,5(mol)

Suy ra : 

V H2 = 0,5.22,4 = 11,2(lít)

m FeCl2 = 0,5.127 = 63,5(gam)

c)

Sau phản ứng: 

mdd = m Fe + mdd HCl - m H2 = 28 + 365 - 0,5.2 = 392(gam)

=> C% FeCl2 = 63,5/392   .100% = 16,2%

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

Ta có: \(n_{Ba\left(OH\right)_2}=\dfrac{34,2}{171}=0,2\left(mol\right)\)

a. PTHH: \(Ba\left(OH\right)_2+Na_2SO_4--->BaSO_4\downarrow+2NaOH\)

b. Theo PT: \(n_{BaSO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)

=> \(m_{BaSO_4}=0,2.233=46,6\left(g\right)\)

c. Theo PT: \(n_{Na_2SO_4}=n_{Ba\left(OH\right)_2}=0,2\left(mol\right)\)

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

=> \(C_{\%_{Na_2SO_4}}=\dfrac{28,4}{240}.100\%=11,83\%\)