Fe2O3+3H2-to>2Fe+3H2O

0,1125---0,3375----0,225 mol

n Fe=0,225 mol

=>m Fe2O3=0,1125.160=18g

=>VH2=0,3375.22,4=7,56l

a,\(n_{Fe}=\dfrac{12,6}{56}=0,225\left(mol\right)\)

PTHH: 3H₂ + Fe₂O₃ ---t^o---> 2Fe + 3H₂O

Mol:   0,3375  0,1125

\(m_{Fe_2O_3}=0,1125.160=18\left(g\right)\)

b, \(V0,3375.22,4=7,56\left(l\right)\) 

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.

a, \(Fe_3O_4+4H_2\underrightarrow{t^o}3Fe+4H_2O\)

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

Theo PT: \(n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{1}{15}\left(mol\right)\Rightarrow m_{Fe_3O_4}=\dfrac{1}{15}.232=\dfrac{232}{15}\left(g\right)\)

c, \(n_{H_2}=\dfrac{4}{3}n_{Fe}=\dfrac{4}{15}\left(mol\right)\Rightarrow V_{H_2}=\dfrac{4}{15}.22,4=\dfrac{448}{75}\left(l\right)\)

d, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(n_{Zn}=n_{H_2}=\dfrac{4}{15}\left(mol\right)\Rightarrow m_{Zn}=\dfrac{4}{15}.65=\dfrac{52}{3}\left(g\right)\)

\(n_{HCl}=2n_{H_2}=\dfrac{8}{15}\left(mol\right)\Rightarrow m_{HCl}=\dfrac{8}{15}.36,5=\dfrac{292}{15}\left(g\right)\)

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) 

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

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