Hóa học

Tham khảo:
 

a)

\(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{11,2}{56}=0,2mol\)

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

  1             3         2          3       ( mol )

0,1           0,3      0,2                 ( mol )

\(m_{Fe_2O_3}=n_{Fe_2O_3}.M_{Fe_2O_3}=0,1.160=16g\)

\(V_{H_2}=n_{H_2}.22,4=0,3.22,4=6,72l\)

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a) \(n_{H_2O}=\dfrac{9}{18}=0,5\left(mol\right)\)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

          0,5<---------------0,5

=> n_H2 = 0,5 (mol)

=> n_HCl = 1 (mol)

Theo ĐLBTKL: m_KL + m_HCl = m_muối + m_H2

=> m_muối = 18,4 + 1.36,5 - 0,5.2 = 53,9 (g)

b) 

PTHH: A + 2HCl --> ACl₂ + H₂

            2B + 6HCl --> 2BCl₃ + 3H₂

=> \(\dfrac{n_{ACl_2}}{n_{BCl_3}}=\dfrac{1}{1}\)

=> \(\dfrac{n_A}{n_B}=\dfrac{1}{1}\)

Có: n_HCl = 2n_A + 3n_B

=> 5n_B = 1

=> n_B = 0,2 (mol) => n_A = 0,2 (mol)

TH1: M_A = 2,4.M_B

Có: n_A.M_A + n_B.M_B = 18,4

=> n_B.2,4.M_B + n_BM_B = 18,4

=> 3,4.n_B.M_B = 18,4

=> M_B = \(27\) (Al)

n_A = 0,2

=> M_A = 65 (Zn)

TH2: M_B = 2,4.M_A

Có: n_A.M_A + n_B.M_B = 18,4

=> n_A.M_A + n_A.2,4.M_A = 18,4

=> 3,4.n_A.M_A = 18,4

=> \(M_A=27\) (Loại do Al hóa trị III)

\(2NaCl+2H_2O\underrightarrow{đp\left(covachngan\right)}2NaOH+Cl_2+H_2\)

tham khảo
NaCl + H2O → NaOH + Cl2 + H2

a) 

\(C\%=\dfrac{25}{25+858}.100\%=2,83\%\)

b) 

\(n_{NaCl}=\dfrac{8,19}{58,5}=0,14\left(mol\right)\)

=> \(C_M=\dfrac{0,14}{0,2}=0,7M\)

Câu 1:

a)

Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)

TH1: Al dư

PTHH: Fe₂O₃ + 2Al --t^o--> Al₂O₃ + 2Fe

                 a--->2a-------->a------->2a

=> B gồm \(\left\{{}\begin{matrix}Al_2O_3:a\left(mol\right)\\Fe:2a\left(mol\right)\\Al:b-2a\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

            2a------------------------>2a

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

       (b-2a)------------------------>1,5(b - 2a)

=> 2a + 1,5b - 3a = 0,1

=> 1,5b - a = 0,1 

Rắn không tan là Fe

=> \(n_{Fe}=2a=\dfrac{13,6}{56}=\dfrac{17}{70}\left(mol\right)\)

=> \(a=\dfrac{17}{140}\left(mol\right)\) => \(b=\dfrac{31}{210}\left(mol\right)\)

Xét \(n_{Al\left(dư\right)}=b-2a=\dfrac{-2}{21}\) (vô lí)

TH2: Fe₂O₃ dư

PTHH: Fe₂O₃ + 2Al --t^o--> Al₂O₃ + 2Fe

            0,5b<---b--------->0,5b---->b

=> B gồm \(\left\{{}\begin{matrix}Al_2O_3:0,5b\left(mol\right)\\Fe:b\left(mol\right)\\Fe_2O_3:a-0,5b\left(mol\right)\end{matrix}\right.\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

             b------------------------>b

=> b = 0,1

Rắn không tan gồm Fe và Fe₂O₃

=> \(56b+160\left(a-0,5b\right)=13,6\)

=> a = 0,1

A gồm \(\left\{{}\begin{matrix}Fe_2O_3:m_{Fe_2O_3}=0,1.160=16\left(g\right)\\Al:m_{Al}=0,1.27=2,7\left(g\right)\end{matrix}\right.\)

B gồm \(\left\{{}\begin{matrix}Al_2O_3:m_{Al_2O_3}=0,05.102=5,1\left(g\right)\\Fe:m_{Fe}=0,1.56=5,6\left(g\right)\\Fe_2O_3:m_{Fe_2O_3}=0,05.160=8\left(g\right)\end{matrix}\right.\)

b) 

Rắn không tan gồm \(\left\{{}\begin{matrix}Fe:0,1\left(mol\right)\\Fe_2O_3:0,05\left(mol\right)\end{matrix}\right.\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

           0,1-->0,2

             Fe₂O₃ + 6HCl --> 2FeCl₃ + 3H₂O

            0,05---->0,3

=> n_HCl = 0,2 + 0,3 = 0,5 (mol)

=> \(V_{ddHCl}=\dfrac{0,5}{0,5}=1\left(l\right)\)

Câu 2:

\(n_{H_2}=\dfrac{2,016}{22,4}=0,09\left(mol\right)\)

\(n_{Al\left(OH\right)_3}=\dfrac{7,8}{78}=0,1\left(mol\right)\)

Gọi số mol Al₂O₃ là x (mol)

PTHH: 2Al + 2NaOH + 2H₂O --> 2NaAlO₂ + 3H₂

         0,06<---------------------0,06<--------0,09

           Al₂O₃ + 2NaOH --> 2NaAlO₂ + H₂O

              x------------------>2x

            2NaAlO₂ + CO₂ + 3H₂O --> Na₂CO₃ + 2Al(OH)₃

               0,1<-------------------------------------0,1

=> 0,06 + 2x = 0,1

=> x = 0,02 (mol)

PTHH: 8Al + 3Fe₃O₄ --t^o--> 4Al₂O₃ + 9Fe

Có \(\dfrac{n_{Al_2O_3}}{n_{Fe}}=\dfrac{4}{9}\)

=> n_Fe = 0,045 (mol)

m_Y = m_X = 11,98 (g)

=> \(m_{Al_2O_3}+m_{Al\left(Y\right)}+m_{Fe}+m_{Fe_3O_4\left(Y\right)}=11,98\)

=> \(m_{Fe_3O_4\left(Y\right)}=5,8\left(g\right)\)

=> \(n_{Fe_3O_4\left(Y\right)}=\dfrac{5,8}{232}=0,025\left(mol\right)\)

Bảo toàn Fe: \(n_{Fe_3O_4\left(bđ\right)}=0,04\left(mol\right)\)

Bảo toàn Al: n_Al(bđ) = 0,1 (mol)

PTHH: 8Al + 3Fe₃O₄ --t^o--> 4Al₂O₃ + 9Fe

Xét tỉ lệ: \(\dfrac{0,1}{8}< \dfrac{0,04}{3}\) => Hiệu suất tính theo Al

PTHH: 8Al + 3Fe₃O₄ --t^o--> 4Al₂O₃ + 9Fe

         0,04<------------------0,02

=> \(H\%=\dfrac{0,04}{0,1}.100\%=40\%\)

Gọi \(\left\{{}\begin{matrix}n_{C_nH_{2n+2}}=a\left(mol\right)\\n_{C_mH_{2m}}=b\left(mol\right)\end{matrix}\right.\)

=> \(a+b=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

\(n_{CO_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

\(n_{H_2O}=\dfrac{30,6}{18}=1,7\left(mol\right)\)

Bảo toàn C: an + bm = 1,5

=> 2an + 2bm = 3 (1)

Bảo toàn H: 2an + 2a + 2bm = 3,4 (2)

(2) - (1) => 2a = 0,4 

=> a = 0,2 (mol)

\(\%V_{C_nH_{2n+2}}=\dfrac{0,2}{0,5}.100\%=40\%\)

\(\%V_{C_mH_{2m}}=100\%-40\%=60\%\)