a ) \(n_{Fe_2O_4}=\frac{23,2}{232}=0,1\) mol

\(Fe_3O_4+4H_2\underrightarrow{t^0}3Fe+4H_2O\)

0,1 -> 0,4 -> 0,3

\(\Rightarrow n_{H_2}=4n_{Fe_3O_4}=0,4\) mol \(\Rightarrow V_{H_2}=0,4.22,4=8,96\) lít

b ) \(n_{Fe}=3n_{Fe_3O_4}=0,3\) mol \(\Rightarrow m_{Fe}=56.0,3=16,8\) gam.

Fe₂O₃+3H₂-to>2Fe+3H₂O

0,1-----0,3-------0,2-----0,3 mol

n Fe₂O₃ =0,1 mol

m Fe=0,2.56=11,2g

VH2=0,3.22,4=6,72l

#yT

Ta có: \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)

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

_____0,3_____0,9___0,6____0,9 (mol)

a, \(m_{Fe}=0,6.56=33,6\left(g\right)\)

b, \(V_{H_2}=0,9.22,4=20,16\left(l\right)\)

c, PT: \(2H_2+O_2\underrightarrow{t^o}2H_2O\)

Theo PT: \(n_{O_2}=\dfrac{1}{2}n_{H_2O}=0,9\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,9.22,4=20,16\left(l\right)\)

tác dụng sắt ,mà ra đồng ??

hình như bn ghi sai r đó phải là:đồng oxit mới phải chứ

a) PTHH:       H₂ + CuO --t⁰--> Cu + H₂O

b) n_CuO = \(\dfrac{m}{M}\) = \(\dfrac{7,2}{80}\) = 0,09 (mol)

n_Cu = n_CuO = 0,09 (mol)

⇒ m_Cu = n x M = 0,09 x 64 = 5,76 (g)

c) n_H2 = n_CuO = 0,09 (mol)

⇒ V_H2 = n x 22,4 = 0,09 x 22,4 = 2,016 (l)

CuO+H2-to>Cu+H2O

0,09----0,09---0,09

n CuO=\(\dfrac{7,2}{80}\)=0,09 mol

=>m Cu=0,09.64=5,76g

=>VH2=0,09.22,4=2,016l

\(n_{CuO}=\dfrac{7,2}{80}=0,09mol\)

\(CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\)

0,09    0,09           0,09              ( mol )

\(m_{Cu}=0,09.64=5,76g\)

\(V_{H_2}=0,09.22,4=2,016l\)

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

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