a;

Theo định luật BTKL ta có:

m_O + m_Fe=m_Fe3O4

mà theo giả thiết thì m_O + m_Fe >m_Fe3O4 nên Fe còn dư

b;

3Fe + 2O₂ \(\rightarrow\)Fe₃O₄

n_Fe3O4=\(\dfrac{16}{232}=0,07\left(mol\right)\)

Theo PTHH ta có:

2n_Fe3O4=n_O2=0,14(mol)

3n_Fe3O4=n_Fe=0,21(mol)

V_O2=22,4.0,14=3,136(lít)

m_{Fe đã Pư}=0,21.56=11,76(g)

m_{Fe dư}=16,8-11,76=5,04(g

A

A

a) \(n_{Fe}=\dfrac{m_{Fe}}{M_{Fe}}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{16}{232}=0,07\left(mol\right)\)

PTHH:

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

(mol) 3..............2...............1

(mol) 0,21........0,14.........0,07

*Tỉ lệ mol:

\(n_{Fe}:n_{Fe_3O_4}=\dfrac{0,3}{3}>\dfrac{0,07}{1}\)

\(\Rightarrow Fe\) dư

b) \(V_{O_2}=n_{O_2}.22,4=0,14.22,4=3,136\left(l\right)\)

n_{Fe dư} = 0,3 - 0,21 = 0,09 (mol)

m_{Fe dư} = n_{Fe dư} . M_{Fe dư} = 0,09 . 56 = 5,04 (g).

Nhấn vào đây

PTHH: \(4Fe+3O_2\rightarrow2Fe_2O_3\)

Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\\n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,3}{4}>\dfrac{0,1}{2}\) \(\Rightarrow\) Sắt còn dư

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=\dfrac{3}{2}n_{Fe_2O_3}=0,15\left(mol\right)\\n_{Fe\left(dư\right)}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe\left(dư\right)}=0,1\cdot56=5,6\left(g\right)\\V_{O_2}=0,15\cdot22,4=3,36\left(l\right)\end{matrix}\right.\)

\(a,3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{16,8}{56}=0,3\left(kmol\right)\\ n_{O_2}=\dfrac{2}{3}.0,3=0,2\left(kmol\right)\\ V_{O_2\left(\text{đ}ktc\right)}=0,2.1000.22,4=4480\left(l\right)\\ n_{Fe_3O_4}=\dfrac{1}{3}.0.3=0,1\left(kmol\right)\\ m_{Fe_3O_4}=232.0,1=23,2\left(kg\right)\)

\(n_{Fe}=\dfrac{6,8}{56}=0,12mol\)

3Fe + 2O₂ \(\underrightarrow{t^o}\) Fe₃O₄

0,12   0,08          0,04  ( mol )

a, \(V_{O_2}=0,08.22,4=1,792l\)

b, m_Fe3O4 = 0,04.232 = 9,28g

\(n_{Fe}=\dfrac{6,8}{56}=\dfrac{17}{140}(mol)\\ PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ a,n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{17}{210}(mol)\\ \Rightarrow V_{O_2}=\dfrac{17}{210}.22,4=1,81(g)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{17}{420}(mol)\\ \Rightarrow m_{Fe_3O_4}=\dfrac{17}{420}.232=9,39(g)\)

\(PTHH:3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

\(Áp.dụng.ĐLBTKL,ta.có:m_{Fe}+m_{O_2}=m_{Fe_3O_4}\\ \Rightarrow m_{O_2}=m_{Fe_3O_4}-m_{Fe}=23,2-16,8=6,4\left(g\right)\\ \Rightarrow n_{O_2}=\dfrac{m}{M}=\dfrac{6,4}{32}=0,2\left(mol\right)\\ \Rightarrow V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

3Fe+2O2->Fe3O4

n_Fe3O4=23,2/232=0,1 mol

=>n_O2=0,1x2=0,2 mol

V_O2=0,2x22,4=4,48 l 

a)\(n_{KMnO_4}=\dfrac{31,6}{158}=0,2\left(m\right)\)

\(PTHH:2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\)

tỉ lệ        :2                  1                 1              1

số mol   :0,2               0,1              0,1           0,1

\(V_{O_2}=0,1.22,4=2,24\left(l\right)\)

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

\(PTHH:3Fe+2O_2\underrightarrow{ }Fe_3O_4\)

theo phương trình ta có tỉ lệ\(\dfrac{0,2}{3}>\dfrac{0,1}{2}\)=>Fe dư

\(PTHH:3Fe+2O_2\xrightarrow[]{}Fe_3O_4\)

tỉ lệ       :3           2         1

số mol  :0,15       0,1      0,05

\(m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\)

\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(n_{O_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

\(\dfrac{0,3}{3}\) < \(\dfrac{0,25}{2}\)                        ( mol )

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

Chất dư là O₂

\(m_{O_2\left(dư\right)}=\left(0,25-0,2\right).32=1,6g\)

\(m_{Fe_3O_4}=0,1.232=23,2g\)

cách tính mol Fe4O4 sao vậy ạ