3fe + 2o2 -nhiệt độ> fe3o4 (1)

nfe=m/M=16.8/56=0.3 mol

nfe3o4 pt =1/3nfe=1/3.0.3=0.1 mol

mfe3o4 pt =n.M=0.1.232=23.2 g

=>hiệu suất = mfe3o4/mfe3o4 pt . 100%=16.8/23.2.100%=72.414%

a)

\(n_{P_2O_5} = \dfrac{42,6}{142} = 0,3(mol)\\ \)

4P      +      5O₂ \(\xrightarrow{t^o}\)         2P₂O₅

0,6.............0,75.................0,3..........(mol)

m_P = 0,6.31 = 18,6(gam)

b)

2KClO₃  \(\xrightarrow{t^o}\)   2KCl     +    3O₂

0,5....................................0,75.....(mol)

\(m_{KClO_3} = 0,5.122,5 = 61,25(gam)\)

c)

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

\(n_{Fe} = \dfrac{16,8}{56} = 0,3(mol)\\ \dfrac{n_{Fe}}{3} = 0,1 < \dfrac{n_{O_2}}{2} = 0,375\)

nên hiệu suất tính theo số mol Fe.

\(n_{Fe\ pư} = 0,3.90\% = 0,27(mol)\\ n_{Fe_3O_4} =\dfrac{1}{3}n_{Fe\ pư} = 0,09(mol)\\ \Rightarrow m_{Fe_3O_4} = 0,09.232 = 20,88(gam)\)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Gọi n H2O = n H2 = a(mol)

Bảo toàn khối lượng:  

20 + 2a = 16,8 + 18a

=> a = 0,2(mol)

n CuO pư = n H2 = 0,2(mol)

Vậy : H = 0,2.80/20   .100% = 80%

\(n_{CuO\left(pư\right)}=a\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(a.............a\)

\(m_{cr}=20-80a+64a=16.8\left(g\right)\)

\(\Leftrightarrow a=0.2\)

\(H\%=\dfrac{0.2\cdot80}{20}\cdot100\%=80\%\)

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

Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,2mol\\n_{Fe_3O_4}=0,1mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_3O_4}=0,1\cdot232=23,2\left(g\right)\\V_{O_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

n \(Fe_3O_4=\dfrac{16,8}{232}0,07\left(mol\right)\) nFe=\(\dfrac{16,8}{56}0,3\left(mol\right)\)

3\(Fe\) + 2 O2 \(\rightarrow\) Fe3O4

3 : 2 :1

0,3 \(\rightarrow\)0,07

H\(\dfrac{0,07.100}{0.3}=23.3\)

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

PTHH : 3Fe + 2O₂ -t^o> Fe₃O₄

             0,3        0,2        0,1

b. \(m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)

c. \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

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

b \(\Rightarrow n_{Fe}=\dfrac{16,8}{56}=0,3mol\) \(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1mol\Rightarrow m_{Fe_3O_4}=0,1\cdot232=2,32g\)

c \(\Rightarrow n_{O_2}=\dfrac{2}{3}n_{Fe}=0,2mol\Rightarrow V_{O_2}=0,2\cdot22,4=4,48l\)

a, \(3Fe+2O_2\rightarrow Fe_3O_4\)

b, \(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(\Rightarrow3Fe+2O_2\rightarrow Fe_3O_{\text{4}}\)

     3    :   2      :   1

     0,3 :  0,2   : 0,1 

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

c,\(\Rightarrow3Fe+2O_2\rightarrow Fe_3O_{\text{4}}\)

        3    :   2      :   1

       0,3 :  0,2   : 0,1 

\(\Rightarrow V_{O_2}=0,2.22,4=4,48\left(lít\right)\)

nFe = 16.8/56 = 0.3 mol 

3Fe + 2O2 -to-> Fe3O4 

0.3___0.2________0.1 

VO2 = 4.48 (l) 

mFe3O4 = 0.1*232 = 23.2 g 

a) 3Fe + 2O₂ --t^o--> Fe₃O₄

Tỉ lệ Fe : O2 = 3:2

b) Áp dụng ĐLBTKL: m_Fe + m_O2 = m_Fe3O4

c) m_Fe + m_O2 = m_Fe3O4

=> m_O2 = 23,2 - 16,8 = 6,4(g)

=> \(n_{O_2}=\dfrac{6,4}{32}=0,2\left(mol\right)=>V_{O_2}=0,2.22,4=4,48\left(l\right)\)

d) Do Fe : O₂ = 3:2

=> Số phân tử O₂ để đốt cháy hết 0,9.10²³ nguyên tử Fe

= \(\dfrac{2}{3}.0,9.10^{23}=0,6.10^{23}\)

BTKL: \(m_{O_2}+m_{Fe}=m_{Fe_3O_4}\)

\(\Rightarrow m_{O_2}=23,2-16,8=6,4(g)\)