\(Fe+2HCl-->FeCl2+H2\)

1) \(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{HCl}=2n_{H2}=0,4\left(mol\right)\)

\(V=V_{HCl}=\frac{0,4}{2}=0,2\left(M\right)\)

2) \(n_{Fe}=n_{H2}=0,2\left(mol\right)\)

\(m_{Fe}=0,2.56=11,2\left(g\right)\)

\(\%m_{Fe}=\frac{11,2}{17,6}.100\%=63,64\%\)

\(\%m_{Cu}=100-63,64=36,36\%\)

3) Ta có

\(m_{Cu}=17,6-11,2=6,4\left(g\right)\)

\(n_{Cu}=\frac{6,4}{64}=0,1\left(mol\right)\)

\(2Fe+6H2SO4-->Fe2\left(SO4\right)3+3SO2+6H2O\)

0,2--------------------------------------------0,3(mol)

\(Cu+2H2SO4-->CuSO4+2H2O+SO2\)

0,1------------------------------------------------------0,1(mol)

Tổng n SO2 = 0,4(mol)

\(V_{O2}=0,4.22,4=8,96\%\)

a,

\(n_{H2}=0,35\left(mol\right)\)

\(n_{HCl}=2n_{H2}=0,7\left(mol\right)\)

\(\Rightarrow m_{dd_{HCl}}=0,7.36,5:15\%=170,33\left(g\right)\)

b,

Gọi a là mol Mg, b là mol Al

\(\Rightarrow24a+27b=7,5\left(1\right)\)

Bảo toàn e: \(2a+3b=0,35.2=0,7\left(2\right)\)

\(\left(1\right)+\left(2\right)\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)

\(\Rightarrow\%_{Mg}=\frac{0,2.24}{7,5}.100\%=64\%\)

\(\Rightarrow\%_{Al}=100\%-64\%=36\%\)

1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2

nH2(1)=3,36/22,4=0.15(mol)

=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)

2.Mg + 2HCl ==> MgCl2 + H2

3.2Al + 6HCl ==> 2AlCl3 + 3H2

4.Fe + 2HCl ==> FeCl2 + H2

=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)

=> nH2(3)=0.1*3/2=0.15(mol)

MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl

AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl

FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl

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

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

nH2 = \(\frac{4,48}{22,4}\)= 0,2 mol

PTHH:
Fe + 2HCl\(\rightarrow\) FeCl2 + H2

FeO + 2HCl \(\rightarrow\) FeCl2 + H2O

\(\rightarrow\) nFe = nH2 = 0,2

\(\rightarrow\)mFe = 0,2.56=11,2 g \(\rightarrow\)mFeO = 18,3 -11,2 = 7,2 g

\(\rightarrow\) nFeO =\(\frac{7,2}{72}\) = 0,1 mol

nHCl = 2 (nFe+nFeO) = 0,6 mol

\(\Rightarrow\) mHCl = 36,5 .0,6 = 21,9

\(\Rightarrow\) C%HCl = \(\frac{21,9}{200}.100\%\) = 43,8%

Bảo toàn khối lượng :

mddsaupứ = mFe + mFeO + mddHCl - mH2

= 18,4 + 200 - 0,4 = 218 g

nFeCl2 = nFe + nFeO = 0,3 mol

mFeCl2 = 127. 0,3 = 38,1 g

C%FeCl2 = \(\frac{38,1}{218}.100\%\) = 17,48%

sai