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

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

a. PTHH: 3NaOH + AlCl₃ ---> Al(OH)₃↓ + 3NaCl (1)

Ta có: \(C_{\%_{NaOH}}=\dfrac{m_{NaOH}}{100}.100\%=12\%\)

=> m_NaOH = 12(g)

=> \(n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\)

Ta lại có: \(C_{\%_{AlCl_3}}=\dfrac{m_{AlCl_3}}{200}.100\%=13,35\%\)

=> \(m_{AlCl_3}=26,7\left(g\right)\)

=> \(n_{AlCl_3}=\dfrac{26,7}{133,5}=0,2\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{3}< \dfrac{0,2}{1}\)

Vậy AlCl₃ dư

Theo PT₍₁₎: \(n_{Al\left(OH\right)_3}=\dfrac{1}{3}.n_{NaOH}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\)

=> \(m_{Al\left(OH\right)_3}=0,1.78=7,8\left(g\right)\)

b. Ta có: \(m_{dd_{NaCl}}=12+200-7,8=204,2\left(g\right)\)

Theo PT₍₁₎: \(n_{NaCl}=n_{NaOH}=0,3\left(mol\right)\)

=> \(m_{NaCl}=0,3.58,5=17,55\left(g\right)\)

=> \(C_{\%_{NaCl}}=\dfrac{17,55}{204,2}.100\%=8,59\%\)

c. PTHH: 2Al(OH)₃ ---t^o---> Al₂O₃ + 3H₂O (2)

Theo PT₍₂₎: \(n_{Al_2O_3}=\dfrac{1}{2}.n_{Al\left(OH\right)_3}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)

=> \(m_{Al_2O_3}=0,05.102=5,1\left(g\right)\)

a)b)c)d) mBaCl2=150.16,64%=24,96g

=>nBaCl2=0,12 mol

mH2SO4=100.14,7%=14,7g=>nH2SO4=0,15mol

     BaCl2       + H2SO4 =>BaSO4    +2HCl

Bđ: 0,12 mol;    0,15 mol

Pứ: 0,12 mol=>0,12 mol=>0,12 mol=>0,24 mol

Dư:                   0,03 mol

Dd ban đầu chứa BaCl2 0,12 mol và H2SO4 0,15 mol

Dd A sau phản ứng chứa HCl 0,24 mol và H2SO4 dư 0,03 mol

mHCl=0,24.36,5=8,76g

mH2SO4=0,03.98=2,94g

Kết tủa B là BaSO4 0,12 mol=>mBaSO4=0,12.233=27,96g

mddA=mddBaCl2+mddH2SO4-mBaSO4

=150+100-27,96=222,04g

C%dd HCl=8,76/222,04.100%=3,945%

C% dd H2SO4=2,94/222,04.100%=1,324%

e) HCl     +NaOH =>NaCl +H2O

0,24 mol=>0,24 mol

H2SO4 +2NaOH =>Na2SO4 + 2H2O

0,03 mol=>0,06 mol

TÔNG nNaOH=0,3 mol

=>V dd NaOH=0,3/2=0,15 lit

a) CuSO4 + 2NaOH --> Cu(OH)2 +Na2SO4(1)

b) nCuSO4=0,03(mol)

nNaOH=0,5(mol)

lập tỉ lệ :

\(\dfrac{0,03}{1}< \dfrac{0,5}{2}\)

=> CuSO4 hết ,NaOH dư => bài toán tính theo CuSO4

theo (1) : nNa2SO4=nCu(OH)2=nCuSO4=0,03(mol)

=>mCu(OH)2=2,94(g)

c) mdd sau pư=25+100-2,94=122,06(g)

C%dd Na2SO4=3,49(%)

theo (1) : nNaOH(pư)=2nCuSO4=0,06(mol)

=>nNaOHdư =0,44(mol)

=>C%dd NaOH dư=14,42(%)

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

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

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

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

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

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=100.14,6\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{HCl\left(pư\right)}=2n_{Zn}=0,2\left(mol\right)\\n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)

Ta có: m _{dd sau pư} = 6,5 + 100 - 0,1.2 = 106,3 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,1.136}{106,3}.100\%\approx12,79\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,2.36,5}{106,3}.100\%\approx6,87\%\end{matrix}\right.\)