a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

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

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,1\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,1.160=16\left(g\right)\)

c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,3\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)

Phương trình hóa học của phản ứng:

Fe2O3 + 3H2 → 2Fe + 3H2O.

Khử 1 mol Fe2O3 cho 2 mol Fe.

x mol Fe2O3 → 0,2 mol.

x = 0,2 : 2 =0,1 mol.

m = 0,1 .160 =16g.

Khử 1 mol Fe2O3 cần 3 mol H2.

Vậy khử 0,1 mol Fe2O3 cần 0,3 mol H2.

V= 0,3 .22.4 = 6,72l.

2) 

nH2 = 6.72/22.4 = 0.3 (mol) 

Fe2O3 + 3H2 -to-> 2Fe + 3H2O 

0.1______0.3______0.2

mFe2O3 = 0.1*160 = 16 (g) 

mFe = 0.2*56 = 11.2 (g) 

3) 

nFe3O4 = 11.6/232 = 0.05 (mol) 

3Fe + 2O2 -to-> Fe3O4 

0.15___0.1______0.05 

mFe = 0.15*56 = 8.4 (g) 

VO2 = 0.1*22.4 = 2.24 (l) 

2KClO3 -to-> 2KCl + 3O2

1/15______________0.1 

mKClO3 = 1/15 * 122.5 = 8.167 (g) 

a)

3H2 + Fe2O3  --t^o--> 2Fe + 3H2O

b) nH2 = 6,72/22,4 = 0,3 mol

Từ pt => nFe3O4 = 0,1 mol

=> mFe3O4 = 0,1. 232 = 23,2 g

1.

a) 3H2 + Fe2O3  --t^o--> 2Fe + 3H2O

b) nH2 = 6,72/22,4 = 0,3 mol

Từ pt => nFe3O4 = 0,1 mol

=> mFe3O4 = 0,1. 232 = 23,2 g

c) nFe = 0,2 mol

=> mFe = 0,2 . 56 = 11,2 (g)

a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

b, \(n_{Fe}=\dfrac{24}{56}=\dfrac{3}{7}\left(mol\right)\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=\dfrac{3}{14}\left(mol\right)\Rightarrow m_{Fe_2O_3}=\dfrac{3}{14}.160=\dfrac{240}{7}\left(g\right)\)

c, Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Fe}=\dfrac{9}{14}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{9}{14}.22,4=14,4\left(l\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{24}{56}\approx0,43\left(mol\right)\\ a.PTHH:Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)  

                     2             3         2          3

                   0,43      0,645   0,45     0,645

\(b.m_{Fe_2O_3}=n.M=0,43.\left(56.2+16.3\right)=68,8\left(g\right)\\ c.V_{H_2}=n.24,79=0,645.24,79=15,98955\left(l\right).\)

a)\(PTHH:Fe_3O_4+4H_2\xrightarrow[]{}3Fe+4H_2O\)

b)\(m_{Fe}=\dfrac{24}{56}=0,4\left(m\right)\)

\(PTHH:Fe_3O_4+4H_2\xrightarrow[]{}3Fe+4H_2O\)

tỉ lệ       :1              4           3         4

số mol  :0,13         0,53       0,4     0,53

\(m_{Fe_3O_4}=0,13.232=30,16\left(g\right)\)

c)\(V_{H_2}=0,53.22,4=11,872\left(l\right)\)

nFe=0,2(mol)

a) PTHH: Fe2O3 + 3 H2 -to-> 2 Fe + 3 H2O

0,1_____________0,3____0,2(mol)

b) mFe2O3=160.0,1=16(g)

c) V(H2,đktc)=0,3.22,4=6,72(l)

nFe = 11.2/56 = 0.2 (mol) 

FeO + H2 -to-> Fe + H2O 

0.2.....0.2...........0.2

mFeO = 0.2*72 = 14.4 (g) 

VH2 = 0.2*22.4 = 4.48 (l) 

PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

a+b) \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2O_3}=0,2\left(mol\right)\\n_{H_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2O_3}=0,2\cdot160=32\left(g\right)\\V_{H_2}=0,6\cdot22,4=13,44\left(l\right)\end{matrix}\right.\)

c) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)

Theo PTHH: \(n_{Zn}=n_{H_2}=0,6\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,6\cdot65=39\left(g\right)\)

a,

nFe = 22,4/56 = 0,4  (mol)

PTHH

Fe₂O₃ + 3H₂ ---t^o----) 2Fe + 3H₂O        (1)

theo phương trình (1) ,ta có:

nFe₂O₃ = 0,4 x 2 / 1 = 0,8 (mol)

mFe₂O₃ = 160 x 0,8 = 128 (g)

b,

theo pt (1)

nH₂ = (0,4 x 3)/2 = 0,6 (mol)

=) VH₂ = 0,6 x 22,4 = 13,44 (L)

c,

PTHH

Zn + H₂SO₄ -------------) ZnSO₄ + H₂                         (2)

Số mol H₂ cần dùng là 0,6 (mol)

Theo PT (2) :

nZn = nH₂    ==) nZn = 0,6 x 65 = 39 (g)

\(n_{Fe}=\dfrac{11.2}{56}=0.2\left(mol\right)\)

\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)

\(0.1.....................0.2\)

\(m_{Fe_2O_3}=0.1\cdot160=16\left(g\right)\)

a, \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

b, \(n_{Fe}=\dfrac{22,4}{56}=0,4\left(mol\right)\)

Theo PT: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe}=0,2\left(mol\right)\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)

c, \(n_{H_2}=\dfrac{3}{2}n_{Fe}=0,6\left(mol\right)\Rightarrow V_{H_2}=0,6.22,4=13,44\left(l\right)\)

d, \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)

a)

$Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O$

b) $n_{Fe} = \dfrac{22,4}{56} = 0,4(mol)$

Theo PTHH : $n_{Fe_2O_3} = \dfrac{1}{2}n_{Fe} = 0,2(mol)$
$m_{Fe_2O_3} = 0,2.160 = 32(gam)$

c) $n_{H_2} = \dfrac{3}{2}n_{Fe} = 0,6(mol)$
$V_{H_2} = 0,6.22,4 = 13,44(lít)$

d) $2H_2 + O_2 \xrightarrow{t^o} 2H_2O$
$V_{O_2} = \dfrac{1}{2}V_{H_2} = 6,72(lít)$
$V_{kk} = 6,72 : 20\% = 33,6(lít)$