Câu 13:

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\\ PTHH:R_2O_3+3H_2\underrightarrow{t^o}2R+3H_2O\\ Theo.pt:n_{R_2O_3}=\dfrac{1}{3}n_{H_2}=\dfrac{1}{3}.0,3=0,1\left(mol\right)\\ M_{R_2O_3}=\dfrac{16}{0,1}=160\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow2R+16.3=160\\ \Leftrightarrow R=56\left(\dfrac{g}{mol}\right)\\ \Leftrightarrow R.là.Fe\\ CTHH:Fe_2O_3\)

Bài 14:

\(n_{H_2}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ PTHH:Fe+H_2SO_{4\left(loãng\right)}\rightarrow FeSO_4+H_2\uparrow\left(1\right)\\ Theo.pt\left(1\right):n_{Fe}=n_{H_2}=0,125\left(mol\right)\\ PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\left(2\right)\\ Theo.pt\left(2\right):n_{Fe_2O_3}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{3}.0,125=\dfrac{1}{24}\left(mol\right)\\ m=m_{Fe_2O_3}=\dfrac{1}{24}.160=\dfrac{20}{3}\left(g\right)\\ n=n_{Fe}=0,125.56=7\left(g\right)\)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ 2Fe+3Cl_2\rightarrow\left(t^o\right)2FeCl_3\\ n_{Cl_2}=\dfrac{3}{2}.0,2=0,3\left(mol\right)\\ n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\\ a,V_{Cl_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,m_{FeCl_3}=162,5.0,2=32,5\left(g\right)\)

a, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(2Fe+3Cl_2\underrightarrow{t^o}2FeCl_3\)

Theo PT: \(n_{Cl_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{Cl_2}=0,3.22,4=6,72\left(l\right)\)

b, \(n_{FeCl_3}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_3}=0,2.162,5=32,5\left(g\right)\)

\(n_{H_2\left(2\right)}=\dfrac{2,8}{22,4}=0,125\left(mol\right)\\ PTHH:Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\left(1\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\left(2\right)\\ n_{Fe}=n_{H_2\left(2\right)}=0,125\left(mol\right)\\ n_{Fe_2O_3}=\dfrac{0,125}{2}=0,0625\left(mol\right)\\ \Rightarrow a=m_{Fe_2O_3}=160.0,0625=10\left(g\right)\\ b=m_{Fe}=0,125.56=7\left(g\right)\)

a pop cs đg onl hem ta

\(n_{FeO}=\dfrac{3.2}{72}=\dfrac{2}{45}\left(mol\right)\)

\(FeO+H_2\underrightarrow{^{^{t^0}}}Fe+H_2O\)

\(\dfrac{2}{45}....\dfrac{2}{45}....\dfrac{2}{45}\)

\(V_{H_2}=\dfrac{2}{45}\cdot22.4=1\left(l\right)\)

\(Fe+\dfrac{3}{2}Cl_2\underrightarrow{^{t^0}}FeCl_3\)

\(\dfrac{2}{45}.............\dfrac{2}{45}\)

\(m_{FeCl_3}=\dfrac{2}{45}\cdot162.5=7.22\left(g\right)\)

\(2Fe+3Cl_2\rightarrow2FeCl_3\)

0,2......0,3..........0,2.....

\(\Rightarrow A=m_{Fe}=11,2g\)

\(\Rightarrow B=m_{FeCl3}=32,5g\)

\(FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

.0,2...........0,6...............0,2........

\(\Rightarrow C=m_{ddNaOH}=480g\)

\(2Fe\left(OH\right)_3\rightarrow Fe_2O_3+3H_2O\)

...0,2..................0,1...........

\(\Rightarrow E=m_{Fe2O3}=16g\)
 

a) \(n_{Fe}=\dfrac{12}{56}=\dfrac{3}{14}\left(mol\right)\)

PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

            \(\dfrac{3}{14}\)---------------------->\(\dfrac{3}{14}\)

\(\Rightarrow V_{H_2}=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

b) \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

PTHH: \(ZnO+H_2\xrightarrow[]{t^o}Zn+H_2O\)

Xét tỉ lệ: \(0,1< \dfrac{3}{14}\Rightarrow H_2\) dư

Theo PT: \(n_{Zn}=n_{ZnO}=0,1\left(mol\right)\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(1mol\)                             \(1mol\)

\(\dfrac{3}{14}mol\)                        \(\dfrac{3}{14}mol\)

\(a)n_{Fe}=\dfrac{m}{M}=\dfrac{12}{56}\approx0,21=\dfrac{3}{14}\left(mol\right)\)

\(V_{H_2}=n.22,4=\dfrac{3}{14}.22,4=4,8\left(l\right)\)

\(b)n_{ZnO}=\dfrac{m}{M}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

\(ZnO+H_2\rightarrow Zn+H_2O\)

\(1mol\)    \(1mol\)    \(1mol\)

\(0,1mol\)  \(0,1mol\)  \(0,1mol\)

\(\text{Ta thấy }H_2\text{ dư,ZnO phản ứng hết.Bài toán tính theo ZnO}\)

\(m_{Zn}=n.M=0,1.65=6,5\left(g\right)\)

a.\(m_{Fe}=8.70\%=5,6g\)

  \(m_{Mg}=8-5,6=2,4g\)

b. \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{5,6}{56}=0,1mol\)

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{2,4}{24}=0,1mol\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

0,1                                     0,1  ( mol )

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

0,1                                       0,1     ( mol )

( chỗ này tính thể tính nhé bạn, mình thấy có chữ đktc )

\(V_{H_2}=n_{H_2}.22,4=\left(0,1+0,1\right).22,4=4,48l\)

c.\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{58,25}{232}=0,24mol\)

\(Fe_3O_4+4H_2\rightarrow\left(t^o\right)3Fe+4H_2O\)

 0,24 >     0,2                                       ( mol )

                 0,2            0,15                      ( mol )

\(m_{Fe}=n_{Fe}.M_{Fe}=0,15.56=8,4g\)

%Fe=70% là %mFe đúng không bạn?

a) m_Fe/8=70% \(\Rightarrow\) m_Fe=5,6 (g) \(\Rightarrow\) m_Mg=2,4 (g).

b) Giả sử kim loại tan hoàn toàn trong dung dịch.

n_{khí hidro}=(2.5,6/56+2.2,4/24)/2=0,2 (mol).

m_{khí hidro}=0,2.2=0,4 (g).

c) Lượng nguyên tử oxi trong Fe₃O₄ ban đầu lớn hơn 1 mol, suy ra lượng khí hidro phản ứng là 0,2 mol.

4H₂(0,2 mol) + Fe₃O₄ \(\underrightarrow{t^o}\) 3Fe(0,15 mol) + 4H₂O.

Khối lượng kim loại thu được là 0,15.56=8,4 (g).