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

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

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 

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

`a)PTHH`

    `Fe + 2HCl -> FeCl_2 + H_2`

`0,125`  `0,25`           `0,125`     `0,125`        `(mol)`

`n_[HCl]=[5/100 .182,5]/[36,5]=0,25(mol)`

`b)m_[Fe]=0,125.56=7(g)`

  `V_[H_2]=0,125.22,4=2,8(l)`

`c)m_[HCl]=0,25.36,5=9,125(g)`

  `m_[FeCl_2]=0,125.127=15,875(g)`

`d)C%_[FeCl_2]=[15,875]/[7+182,5-0,125.2] .100~~8,39%`

Fe+2HCl->FeCl₂+H₂

0,125--0,25---0,125-0,125

m _HCl=9,125 g=>n _HCl=\(\dfrac{9,125}{26,5}\)=0,25 mol

=>m Fe=0,125.56=7g

=>V_H2=0,125.22,4=2,8l

=>C%_FeCl2=\(\dfrac{0,125.127}{7+182,5-0,25}\).100=8,388%

\(a) Fe + 2HCl \to FeCl_2\\ b) n_{HCl} = \dfrac{182,5.5\%}{36,5} = 0,25(mol)\\ n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{1}{2}n_{HCl} = 0,125(mol)\\ \Rightarrow m_{Fe} = 0,125.56 = 7(gam) ; V = 0,125.22,4 = 2,8(lít)\\ c) m_{dd\ sau\ phản\ ứng} = m_{Fe} + m_{dd\ HCl} - m_{H_2} = 7 + 182,5 - 0,125.2 = 189,25(gam)\\ C\%_{FeCl_2} = \dfrac{0,125.127}{189,25}.100\% = 8,39\%\)

Fe+2HCl->Fecl2+H2

0,15---0,3-----------0,15

n Fe=0,15 mol

=>VH2=0,15.22,4=3,36l

2H2+O2-to>2H2O

0,15---------------0,15

n O2=0,2 mol

=>O2 dư

=>m H2O=0,15.18=2,7g

a)

\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)

b)

\(n_{CuO} = n_{Cu} = \dfrac{6,4}{64} = 0,1(mol)\\ \Rightarrow m_{CuO} = 0,1.80 = 8(gam)\)

c)

\(n_{O_2} = \dfrac{1}{2}n_{Cu} = 0,05(mol)\\ 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ m_{KMnO_4} = 2n_{O_2} = 0,05.2 = 0,1.158 = 15,8(gam)\)

d)

\(V_{không\ khí} = 5V_{O_2} = 0,05.22,4.5 = 5,6(lít)\)

Ta co pthh

3Fe + 2O2-^to\(\rightarrow\) Fe3O4

Theo de bai ta co

nFe=\(\dfrac{33,6}{56}=0,6mol\)

Theo pthh

nO2=\(\dfrac{2}{3}nFe=\dfrac{2}{3}.0,6=0,4mol\)

\(\Rightarrow VO2_{\left(dktc\right)}\)=0,4.22,4=8,96 l

Theo pthh

nFe3O4=\(\dfrac{1}{3}nFe=\dfrac{1}{3}.0,6=0,2mol\)

\(\Rightarrow mFe3O4=0,2.232=46,4g\)