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)

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

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

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

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

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

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

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

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

Ta có: \(m_{CuSO_4}=\dfrac{10\%.160}{100\%}=16\left(g\right)\)

=> \(n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\)

Ta lại có: \(m_{NaOH}=\dfrac{5\%.240}{100\%}=12\left(g\right)\)

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

a. PTHH: \(CuSO_4+2NaOH--->Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

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

Vậy NaOH dư, CuSO₄ hết.

Theo PT: \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\)

=> \(m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

b. Ta có: \(m_{dd_{Na_2SO_4}}=240+16-9,8=246,2\left(g\right)\)

Ta có: \(m_{Na_2SO_4}=0,1.142=14,2\left(g\right)\)

=> \(C_{\%_{Na_2SO_4}}=\dfrac{14,2}{246,2}.100\%=5,77\%\)

Tiếp: 

Theo PT: \(n_{NaOH}=2.n_{CuSO_4}=2.0,1=0,2\left(mol\right)\)

=> \(m_{NaOH_{PỨ}}=0,2.40=8\left(g\right)\)

=> \(m_{NaOH_{dư}}=12-8=4\left(g\right)\)

=> \(C_{\%_{NaOH_{dư}}}=\dfrac{4}{240}.100\%=1,7\%\)

\(a,m_{CuSO_4}=160.10\%=16\left(g\right)\\ n_{CuSO_4}=\dfrac{16}{160}=0,1\left(mol\right)\\ b,m_{NaOH}=240.5\%=12\left(g\right)\\ n_{NaOH}=\dfrac{12}{40}=0,3\left(mol\right)\\ PTHH:2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\\ Vì:\dfrac{n_{NaOH\left(đề\right)}}{n_{NaOH\left(PTHH\right)}}=\dfrac{0,3}{2}>\dfrac{n_{CuSO_4\left(đề\right)}}{n_{CuSO_4\left(PTHH\right)}}=\dfrac{0,1}{1}\\ \Rightarrow CuSO_4hết,NaOHdư\)

Vậy:

 \(n_{Cu\left(OH\right)_2}=n_{Na_2SO_4}=n_{CuSO_4}=0,1\left(mol\right)\\ \Rightarrow m_{\downarrow}=m_{Cu\left(OH\right)_2}=0,1.98=9,8\left(g\right)\)

\(b,m_{ddB}=m_{ddCuSO_4}+m_{ddNaOH}-m_{\downarrow}=160+240-9,8=390,2\left(g\right)\)

\(n_{NaOH\left(dư\right)}=0,3-0,1.2=0,1\left(mol\right)\)

\(C\%_{ddNa_2SO_4}=\dfrac{0,1.142}{390,2}.100\approx3,639\%\\ C\%_{ddNaOH\left(dư\right)}=\dfrac{0,1.40}{390,2}.100\approx1,025\%\)

\(a,PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \left\{{}\begin{matrix}m_{H_2SO_4}=\dfrac{300\cdot9,8\%}{100\%}=29,4\left(g\right)\\m_{BaCl_2}=\dfrac{200\cdot26\%}{100\%}=52\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{H_2SO_4}=\dfrac{29,4}{98}=0,3\left(mol\right)\\n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\end{matrix}\right.\)

Vì \(\dfrac{n_{H_2SO_4}}{1}>\dfrac{n_{BaCl_2}}{1}\) nên H₂SO₄ dư

\(\Rightarrow n_{BaSO_4}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow a=m_{BaSO_4}=0,25\cdot233=58,25\left(g\right)\\ b,n_{HCl}=n_{BaCl_2}=0,25\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,25\cdot36,5=9,125\left(g\right)\\ m_{dd_{HCl}}=300+200-58,25=441,75\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{9,125}{441,75}\cdot100\%\approx2,07\%\)