\(a,PTHH:CuCl_2+2KOH\rightarrow Cu\left(OH\right)_2+2KCl\\ ...0,2......0,4.......0,2........0,4\left(mol\right)\\ b,n_{CuCl_2}=\dfrac{27}{135}=0,2\left(mol\right)\\ m_{Cu\left(OH\right)_2}=0,2\cdot98=19,6\left(g\right)\\ c,m_{KOH}=0,4\cdot56=22,4\left(g\right)\\ m_{dd_{KOH}}=\dfrac{22,4\cdot100\%}{20\%}=112\left(g\right)\\ m_{dd_{KCl}}=m_{CuCl_2}+m_{dd_{KOH}}-m_{Cu\left(OH\right)_2}=27+112-19,6=119,4\left(g\right)\)

\(d,C\%_{dd_{KCl}}=\dfrac{74,5\cdot0,4}{119,4}\cdot100\%\approx24,96\%\)

Câu 3 : 

\(m_{ct}=\dfrac{10.80}{100}=8\left(g\right)\)

\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)

a) Hiện tượng : Xuất hiện kết tủa trắng

Pt : \(2NaOH+MgSO_4\rightarrow Na_2SO_4+Mg\left(OH\right)_2|\)

           2                1                  1               1

        0,2                0,1               0,1            0,1

\(n_{Mg\left(OH\right)2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{Mg\left(OH\right)2}=0,1.58=5,8\left(g\right)\)

b) \(n_{MgSO4}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

 \(m_{MgSO4}=0,1.120=12\left(g\right)\)

\(m_{ddMgSO}=\dfrac{12.100}{10}=120\left(g\right)\)

c) \(n_{Na2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

⇒ \(m_{Na2SO4}=0,1.142=14,2\left(g\right)\)

\(m_{ddspu}=80+120-5,8=194,2\left(g\right)\)

\(C_{Na2SO4}=\dfrac{14,2.100}{194,2}=7,31\)⁰/₀

 Chúc bạn học tốt

Câu 4 : 

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)

         1         1               1           1

        0,2      0,2

\(n_{H2SO4}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{19,6.100}{20}=98\left(g\right)\)

\(V_{ddH2SO4}=\dfrac{98}{1,2}\simeq81,67\left(ml\right)\)

 Chúc bạn học tốt

a, \(CuCl_2+2NaOH\rightarrow2NaCl+Cu\left(OH\right)_2\)

b, \(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

Theo PT: \(n_{CuCl_2}=n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,125\left(mol\right)\)

\(\Rightarrow m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

c, \(C_{M_{CuCl_2}}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

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

                   0,125<-----0,25---------------------->0,125

b) \(m_{Kt}=m_{Cu\left(OH\right)2}=0,125.98=12,25\left(g\right)\)

c) \(C_{MCuCl2}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(n_{NaOH}=\dfrac{10}{40}=0,25\left(mol\right)\)

PTHH:

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

0,125       0,25                0,125         0,25

\(m_{Cu\left(OH\right)_2}=0,125.98=12,25\left(g\right)\)

\(C_{M\left(CuCl_2\right)}=\dfrac{0,125}{0,1}=1,25\left(M\right)\)

\(m_{Fe_2\left(SO_4\right)_3}=\dfrac{200\cdot20}{100}=40\left(g\right)\Rightarrow n=0,1mol\)

\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)

0,1                    0,6              0,2                   0,3

a)\(m_{NaOH}=0,6\cdot40=24\left(g\right)\)

b)\(m_{Fe\left(OH\right)_3}=0,2\cdot107=21,4\left(g\right)\)

c)\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)

   \(m_{ddsau}=200+24-21,4=202,6\left(g\right)\)

   \(\Rightarrow C\%=\dfrac{42,6}{202,6}\cdot100\%=21,03\%\)

Ta có: \(C_{\%_{KOH}}=\dfrac{m_{KOH}}{112}.100\%=56\%\)

=> m_KOH = 62,72(g)

=> \(n_{KOH}=\dfrac{62,72}{56}=1,12\left(mol\right)\)

a. PTHH: 2KOH + MgCl₂ ---> Mg(OH)₂↓ + 2KCl

Theo PT: \(n_{Mg\left(OH\right)_2}=\dfrac{1}{2}.n_{KOH}=\dfrac{1}{2}.1,12=0,56\left(mol\right)\)

=> \(m_{Mg\left(OH\right)_2}=0,56.58=32,48\left(g\right)\)

b. Theo PT: \(n_{MgCl_2}=n_{Mg\left(OH\right)_2}=0,56\left(mol\right)\)

=> \(m_{MgCl_2}=0,56.95=53,2\left(g\right)\)

=> \(C_{\%_{MgCl_2}}=\dfrac{53,2}{200}.100\%=26,6\%\)

\(n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂SO₄ --> Fe₂(SO₄)₃ + 3H₂O

______0,05------>0,15--------->0,05

=> m_H2SO4 = 0,15.98 = 14,7(g)

=> \(C\%\left(H_2SO_4\right)=\dfrac{14,7}{100}.100\%=14,7\%\)

\(C\%\left(Fe_2\left(SO_4\right)_3\right)=\dfrac{0,05.400}{8+100}.100\%=18,52\%\)

PTHH: Fe₂(SO₄)₃ + 6NaOH --> 2Fe(OH)₃\(\downarrow\) + 3Na₂SO₄

________0,05----------------------->0,1

=> mFe(OH)₃ = 0,1.107=10,7(g)

a/

\(n_{Na_2O}=\dfrac{9,3}{62}=0,15\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,15 0,3 (mol)

\(m_{NaOH}=0,3.40=12\left(g\right)\)

\(m_A=90,7+9,3=100\left(g\right)\)

\(C\%_{NaOH}=\dfrac{12}{100}.100\%=12\%\)

b/

m\(_{FeSO_4}=\dfrac{16.200}{100}=32\left(g\right)\)

\(\rightarrow m_{FeSO_4}=\dfrac{32}{152}=\dfrac{4}{19}\left(mol\right)\)

\(2NaOH+FeSO_4\rightarrow Na_2SO_4+Fe\left(OH\right)_2\downarrow\)

bđ: 0,3 \(\dfrac{4}{19}\) 0 0 (mol)

pư: 0,3 0,15 0,15 0,15 (mol)

dư: 0 \(\dfrac{23}{380}\) (mol)

\(m_{Fe\left(OH\right)_2}=0,15.90=13,5\left(g\right)\)

\(m_C=100+200-13,5=286,5\left(g\right)\)

\(m_{Na_2SO_4}=0,15.142=21,3\left(g\right)\)

\(\rightarrow C\%_{Na_2SO_4}=\dfrac{21,3}{286,5}.100\%\approx7,4\%\)

\(m_{FeSO_4\left(dư\right)}=\dfrac{23}{380}.152=9,2\left(g\right)\)

\(\rightarrow C\%_{FeSO_4\left(dư\right)}=\dfrac{9,2}{286,5}.100\%\approx3,2\%\)