\(n_{O_2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: 2KMnO₄ \(\underrightarrow{t^o}\) K₂MnO₄ + MnO₂ + O₂

3 <--------------------------------------------- 1,5 (mol)

->\(m_{KMnO_4}=3.158:80\%=592,5\left(g\right)\)

\(n_{O_2}=\frac{33,6}{22,4}=1,5\left(mol\right)\)

PTHH: \(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\)

Theo PTHH: \(n_{O_2}:n_{KMnO_4}=1:2\)

\(\Rightarrow n_{KMnO_4}=n_{O_2}.2=1,5.2=3\left(mol\right)\)

\(\Rightarrow m_{KMnO_4}=\frac{3.158}{80\%}=592,5\left(mol\right)\)

\(a)\)

\(Fe_2O_3+3H_2-t^o-> 2Fe+3H_2O\)\((1)\)

\(CuO+H_2-t^o->Cu+H_2O\) \((2)\)

Gọi x là nFe2O3, y là nCuO

Theo đề, \( mCuO:mFe2O3=1:2\)

\(=> mFe_2O_3=2mCuO\)

\(<=> 160x- 2.80y =0\)

\(<=> 160x-160y=0\) \((I)\)

a)

\(n_{KMnO_4\left(bđ\right)}=\frac{126,4}{158}=0,8\left(mol\right)\)

\(n_{KMnO_4\left(pư\right)}=\frac{0,8.80}{100}=0,64\left(mol\right)\)

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

________0,64---------->0,32-------->0,32---->0,32_____(mol)

=> \(m_{K_2MnO_4}=0,32.197=63,04\left(g\right)\)

b) \(m_{MnO_2}=0,32.87=27,84\left(g\right)\)

c) \(V_{O_2}=0,32.22,4=7,168\left(l\right)\)

d) \(m_{rắn}=m_{KMnO_4}\) (không pư) + \(m_{MnO_2}+m_{K_2MnO_4}\)

= \(158\left(0,8-0,64\right)+27,84+63,04\)

= 116,16 (g)

\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)\)

\(2Fe+\dfrac{3}{2}O_2\underrightarrow{t^0}Fe_2O_3\)

\(0.1.......0.075.....0.05\)

\(V_{O_2}=0.075\cdot22.4=1.68\left(l\right)\)

\(m_{Fe_2O_3}=0.05\cdot160=8\left(g\right)\)

PTHH: \(4Fe+3O_2\underrightarrow{t^o}2Fe_2O_3\)

Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{O_2}=0,075\left(mol\right)\\n_{Fe_2O_3}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,075\cdot22,4=1,68\left(l\right)\\m_{Fe_2O_3}=0,05\cdot160=8\left(g\right)\end{matrix}\right.\)

$1) 3Fe + 2O_2 \xrightarrow{t^o} Fe_3O_4$

$2) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5$
$3) 2Na + S \xrightarrow{t^o} Na_2S$
$4) 2KClO_3 \xrightarrow{t^o} 2KCl +3 O_2$
$5) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$6) CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
$7) 2H_2 + O_2 \xrightarrow{t^o} 2H_2O$

1,2,3,7 : Phản ứng hóa hợp

4,5 : Phản ứng phân hủy

6 : Phản ứng thế

1. 3Fe + 2O₂ ---> Fe₃O₄

2. 4P + 5O₂ ---> 2P₂O₅

4. 2KClO₃ ---> 2KCl + 3O₂

5. 2KMnO₄ ---> K₂MnO₄ + MnO₂ + O₂

6. H₂ + CuO ---> Cu + H₂O

7. 4H₂ + O₂ ---> 2H₂O

Fe + 2H+ → Fe2+ + H2↑

Fe2O3 + 6H+ → 2Fe3+ + 3H2O

25ml dd phản ứng với 12,5ml dd KMnO4 0,096M

→ 500ml dd phản ứng với
\(\frac{500.12,5}{25}\) = 250ml dd KMnO4 0,096M

nKMnO4 phản ứng với 500ml dd = 0,096 . 0,25 = 0,024 mol

Fe+2 → Fe+3 + 1e M+7 + 5e → Mn+2

0,12 ← 0,12 0,12←0,024

(Bảo toàn e)

nFe = nFe2+ = 0,12 mol

\(\%m_{Fe}=\frac{0,12.56}{7,81}.100\%=93,59\%\)

C1:

1/

Oxit axit:

- SO3: lưu huỳnh trioxit

- N2O5: đinito pentaoxit

- SiO2 : silic đioxit

Oxit bazo:

- Na2O: natri oxit

- Fe2O3: Sắt (III) oxit

- BaO: bari oxit

2/ Các oxit tác dụng được với nước:

Na2O + H2O --> 2NaOH

SO3 + H2O --> H2SO4

N2O5 + H2O --> 2HNO3

BaO + H2O --> Ba(OH)2

\(n_{H_2}=\frac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH Fe₂O₃ + 3H₂ \(\underrightarrow{t^o}\) 2Fe + 3H₂O

0,4 \(\rightarrow\) \(\frac{4}{15}\) (mol)

m_Fe(tđ) = \(\frac{4}{15}\). 56.90% = 13,44(g)