CHƯƠNG VI: DUNG DỊCH

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ n_{H_2SO_4}=\dfrac{117,6.25\%}{98}=0,3\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\LTL:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow H_2SO_4dư\\ n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ \Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\\DungdịchB:MgSO_4;H_2SO_4dư\\ m_{ddsaupu}=4,8+117,6-0,2.2=122\left(g\right)\\ n_{H_2SO_4dư}=0,3-0,2=0,1\left(mol\right)\\ n_{MgSO_4}=n_{Mg}=0,1\left(mol\right)\\ C\%_{H_2SO_4dư}=\dfrac{0,1.98}{122}.100=8,03\%\\ C\%_{MgSO_4}=\dfrac{0,2.120}{122}.100=19,67\% \)

B1 : nFe = 11,2 /56 = 0,2 (mol)

Fe+ 2HCl -- . FeCl2 + H2

mFeCl2 = 0,2.127 = 25,4 (g)

VH2 = 0,2 .22,4 = 4,48 (l)

mHCl = 0,4.36,5 = 14,6(g)

C%\(_{ddHCl}=\dfrac{ }{ }\)\(\dfrac{14,6.100}{280}=5,2\%\)

C2 :

2Al + 3H2SO4 -- > Al2(SO4)3 + 3H2

nH2 = 17,92/22,4 = 0,8 (mol)

mAl = (2/3.0,8 ) .27 = 14,4 (g)

mAl2(SO4)3 = (1/3 . 0,8 ) . 342 = 91,2 (g)

mH2SO4 = 0,8 . 98 = 78,4 (g)

\(C\%_{ddH_2SO_4}=\dfrac{78,4.100}{120}=65,33\%\)

nNa = 4,6 / 23 = 0,2 (mol)

Na + 2H2O -- > NaOH + H2

mH2 = 0,2.2 =0,4 (g)

dd = mNa + mddH2O - mH2 = 4,6 + 200 - 0,4 = 204,2(g)

mNaOH = 0,2 . 40 = 8(g)

=> \(C\%_{ddA}=\dfrac{8.100}{204,2}=3,9\%\)

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

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

            0,1-->0,2

=> \(C_{M\left(dd.HCl\right)}=\dfrac{0,2}{0,2}=1M\)

`n_(Zn) = (6,5)/65 = 0,1 (mol)`

`PTHH: ZN + 2HCl --> ZnCl_2 + H_2`

`0,1 --> 0,2`

`=> C_(M(dd.HCl) = (0,2)/(0,2) = 1M`

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

PTHH: Zn + H₂SO₄ ---> ZnSO₄ + H₂

         0,2--->0,2--------->0,2------>0,2

\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ m_{H_2SO_4}=\dfrac{0,2.98}{20\%}=98\left(g\right)\\ \rightarrow V_{ddH_2SO_4}=\dfrac{98}{1,14}=86\left(ml\right)=0,086\left(l\right)\\ \rightarrow C_{M\left(H_2SO_4\right)}=\dfrac{0,2}{0,086}=2,33M\)

`Mg + 2HCl -> MgCl_2 + H_2`

`0,15`  `0,3`                           `0,15`       `(mol)`

`n_[Mg]=[3,6]/24=0,15(mol)`

`a)V_[H_2]=0,15.22,4=3,36(l)`

`b)m_[HCl]=0,3.36,5=10,95(g)`

`c)`

`H_2 + CuO` $\xrightarrow{t^o}$ `Cu + H_2 O`

`0,15`                  `0,15`                 `(mol)`

  `=>m_[Cu]=0,15.64=9,6(g)`

\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)

PTHH: Mg + 2HCl ---> MgCl₂ + H₂

          0,15->0,3------------------>0,15

          CuO + H₂ --t^o--> Cu + H₂O

                    0,15------>0,15

=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\\ m_{HCl}=0,3.36,5=10,95\left(g\right)\\ m_{Cu}=0,15.64=9,6\left(g\right)\)

a)

C% muối =\(\dfrac{25}{25+35+190}100=10\%\)

C% đường =\(\dfrac{35}{25+35+190}.100=14\%\)

b)

200ml=0,2, x(ml)=0,001x(l)

nNaOH=0,2×3=0,6(mol)

→CMNaOH 1,5M=0,6\0,2+0,001x=1,5M

       →x=200

\(C\%_{FeBr_2.bão.hoà}=\dfrac{115}{115+100}.100\%=53,5\%\\ \rightarrow m_{FeBr_2.6H_2O}=53,5\%.516=276\left(g\right)\)

\(n_{K_2O}=\dfrac{21,15}{94}=0,225\left(mol\right)\\ pthh:K_2O+H_2O\rightarrow KOH+\dfrac{1}{2}H_2\) 
         0,225                   0,225        0,1125 
\(m_{\text{dd}}=21,15+178,85-\left(0,1125.2\right)=199,775g\\ C\%=\dfrac{0,225.56}{199,775}.100\%=6,3\%\) 
c)
\(m_{\text{dd}}=21,15+\left(50+178,85\right)-\left(0,1125.2\right)=249,775g\\ C\%=\dfrac{0,225.56}{249,775}.100\%=5\%\)

\(m_{FeCl_2}=0,5.127=63,5g\)

\(C\%_{FeCl_2}=\dfrac{63,5}{300}.100\%=21,16\%\)

\(m_{FeCl_2}=n.M=0,5.127=63,5\left(g\right)\)

\(C_{\%_{ddFeCl_2}}=\dfrac{m_{FeCl_2}}{m_{ddFeCl_2}}.100\%=\dfrac{63,5}{300}.100\%=21,2\%\)