a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

PTHH: 2CH₃COOH + Mg --> (CH₃COO)₂Mg + H₂

                  0,1<----------------------0,05------->0,05

=> V_H2 = 0,05.22,4 = 1,12 (l)

b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)

a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)

Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(mHCl=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)

\(nHCl=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH:

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

 2         6              2            3    (mol)

0,1       0,3         0,1           0,15   (mol)

LTL : 0,1 / 2 < 0,4/6

=> Al đủ , HCl dư

1. \(VH_2=0,15.22,4=3,36\left(l\right)\)

2. \(mH_2=0,15.2=0,3\left(g\right)\)

mdd = mAl + mddHCl - mH2 = 2,7 + 200 - 0,3 = 202,4 (g)

\(mH_2SO_{4\left(dưsaupứ\right)}=0,1.98=9,8\left(g\right)\)

\(mAlCl_2=0,1.98=9,8\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{9,8.100}{202,4}=4,84\%\)

\(C\%_{AlCl_2}=\dfrac{9,8.100}{202,4}=4,84\%\)

Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

____0,5____________0,5 (mol)

a, \(m_{CuCl_2}=0,5.135=67,5\left(g\right)\)

b, Có: m _{dd sau pư} = m_CuO + m _{dd HCl} = 40 + 200 = 240 (g)

\(\Rightarrow C\%_{CuCl_2}=\dfrac{67,5}{240}.100\%=28,125\%\)

Bạn tham khảo nhé!

\(n_{CH_3COOH}=\dfrac{100.12\%}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaHCO₃ --> CH₃COONa + CO₂ + H₂O

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

=> \(m_{dd.NaHCO_3.8,4\%}=\dfrac{0,2.84.100}{8,4}=200\left(g\right)\)

m_{dd sau pư} = 100 + 200 - 0,2.44 = 291,2 (g)

\(m_{CH_3COONa}=0,2.82=16,4\left(g\right)\)

=> \(C\%_{CH_3COONa}=\dfrac{16,4}{291,2}.100\%=5,632\%\)

Bài 4 : 

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

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

            1          2              1          1

          0,2        0,4           0,2        0,2

b) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)

c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

d) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

Bài 3 : 

\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)

            1            1                1           1

          0,5          0,5            0,5          0,5

b) \(n_{H2}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(V_{H2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

c) \(n_{H2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{H2SO4}=0,5.98=49\left(g\right)\)

\(C_{ddH2SO4}=\dfrac{49.100}{200}=24,5\)⁰/₀

d) \(n_{MgSO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{MgSO4}=0,5.120=60\left(g\right)\)

\(m_{ddspu}=12+200-\left(0,5.2\right)=211\left(g\right)\)

\(C_{MgSO4}=\dfrac{60.100}{211}=28,44\)⁰/₀

 Chúc bạn học tốt

Mg+H2SO4->MgSO4+H2

0,5----0,5--------0,5--------0,5

n Mg=12\24=0,5 mol

=>VH2=0,5.22,4=11,2l

=>m H2SO4=0,5.98=49 g

=>C% =49\200.100=24,5g

=>m dd=12+200-0,5.2=211g

=>C%=0,4.120\211.100=22,74885%

Bài 6:

\(n_{Fe\left(OH\right)_3}=\dfrac{21,4}{107}=0,2\left(mol\right)\)

PT: \(Fe\left(OH\right)_3+3HCl\rightarrow FeCl_3+3H_2O\)

_______0,2________0,6______0,2 (mol)

a, \(C\%_{HCl}=\dfrac{0,6.36,5}{200}.100\%=10,95\%\)

b, \(C\%_{FeCl_3}=\dfrac{0,2.162,5}{21,4+200}.100\%\approx14,68\%\)

Bài 7:

\(m_{H_2SO_4}=100.9,8\%=9,8\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{9,8}{98}=0,1\left(mol\right)\)

PT: \(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\)

______0,1______0,1_______0,1 (mol)

a, \(m_{ZnO}=0,1.81=8,1\left(g\right)\)

b, \(C\%_{ZnSO_4}=\dfrac{0,1.161}{8,1+100}.100\%\approx14,89\%\)

`a)PTHH:`

  `Zn + 2HCl -> ZnCl_2 + H_2`

`0,02`                       `0,02`    `0,02`        `(mol)`

`n_[Zn]=[1,3]/65=0,02(mol)`

`b)V_[H_2]=0,02.22,4=0,448(l)`

`c)C%_[ZnCl_2]=[0,02.136]/[1,3+50-0,02.2].100~~5,31%`

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

0,02   0,04         0,02       0,02  ( mol )

\(V_{H_2}=0,02.22,4=0,448\left(l\right)\)

\(C\%_{ZnCl_2}=\dfrac{0,02.136}{1,3+50-0,02.2}.100=5,3\%\)

\(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) 
         0,02                 0,02           0,02 
\(V_{H_2}=0,02.22,4=0,448\left(l\right)\\ m_{\text{dd}}=1,3+50-\left(0,02.2\right)=51,26\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,02.136}{51,26}.100\%=5,3\%\)