\(m_{CuO}=50.20\%=10\left(g\right)\)

\(n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\)

\(m_{Fe_2O_3}=50-10=40\left(g\right)\)

\(n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\)

PTHH :

  \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

0,125   0,125   0,125 

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

0,25        0,75    0,5 

\(a,V_{H_2}=\left(0,75+0,125\right).22,4=19,6\left(l\right)\)

\(b,m_{Cu}=0,125.64=8\left(g\right)\)

\(m_{Fe}=0,5.56=28\left(g\right)\)

\(m_{CuO}=40.20\%=8\left(g\right)\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{40-8}{160}=0,2\left(mol\right)\)

PTHH:

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

 0,1      0,1

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

0,2          0,6 

\(V_{H_2}=\left(0,1+0,6\right).22,4=15,68\left(l\right)\)

_ \(m_{Fe_2O_3}=0,8.50=40\left(g\right)\) \(\Rightarrow m_{CuO}=50-40=10\left(g\right)\)

_ \(n_{Fe_2O_3}=\dfrac{40}{160}=0,25mol\); \(n_{CuO}=\dfrac{10}{80}=0,125mol\)

PTHH: \(3H_2+Fe_2O_3\rightarrow2Fe+3H_2O\)

_____0,75mol_0,25mol

\(H_2+CuO\rightarrow Cu+H_2O\)

0,125__0,125 (mol)

\(\Rightarrow V_{H_2}=\left(0,75+0,125\right)22,4=19,6l\)

1.

\(M_{CuO}\)= 64 + 16 = 80 (g)

\(M_{Fe_2O_3}\)= 56 .2+16.3 = 160 (g)

Theo đề bài cho:

\(m_{Fe_2O_3}=\frac{80}{100}\times50=40\left(g\right)\)

\(\rightarrow n_{Fe_2O_3}=\frac{40}{160}=0,25\left(mol\right)\)

\(m_{CuO}=\frac{20}{100}\times50=10\left(g\right)\)

\(\rightarrow n_{CuO}=\frac{10}{80}=0,125\left(mol\right)\)

Phương trình hóa học:

\(CuO+H_2\rightarrow Cu+H_2O\) (1)

0,125mol - 0,125 mol

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\) (2)

0,25mol- 0,75 mol

\(\Sigma n_{khí}H_2\)cần dùng: 0,125 + 0,75=0,875 mol

\(\Rightarrow V_{H_2}=0,875\times22,4=19,6l\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

PTHH: Fe₂O₃ + 3H₂ --to--> 2Fe + 3H₂O

             0,1         0,3              0,2

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

\(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: 3Fe + 2O₂ --to--> Fe₃O₄

LTL: \(\dfrac{0,2}{3}>\dfrac{0,1}{2}\rightarrow\) Fe dư

Theo pthh: \(n_{Fe\left(pư\right)}=\dfrac{3}{2}n_{O_2}=\dfrac{3}{2}.0,1=0,15\left(mol\right)\)

\(\rightarrow m_{Fe\left(dư\right)}=\left(0,2-0,15\right).56=2,8\left(g\right)\)

a.\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1mol\)

\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)

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

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

b.\(n_{O_2}=\dfrac{2,24}{22,4}=0,1mol\)

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)

\(\dfrac{0,2}{3}\) > \(\dfrac{0,1}{2}\)                             ( mol )

 0,15           0,1                         ( mol )

Chất dư là Fe

\(m_{Fe\left(dư\right)}=\left(0,2-0,15\right).56=2,8g\)

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

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

n Fe2O3=0,1 mol

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

b) 

3Fe+2O2-to>Fe3O4

0,15---0,1

n O2=0,1 mol

=>Fe dư :

=>m Fe du =0,05.56=2,8g

\(\left\{{}\begin{matrix}m_{CuO}=50.20\%=10\left(g\right)\\m_{Fe_2O_3}=50-10=40\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{CuO}=\dfrac{10}{80}=0,125\left(mol\right)\\n_{Fe_2O_3}=\dfrac{40}{160}=0,25\left(mol\right)\end{matrix}\right.\)

PTHH:
\(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)

0,125->0,125

\(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2\)

0,25--->0,75

\(\Rightarrow V_{H_2}=\left(0,75+0,125\right).22,4=18,2\left(l\right)\)

\(m_{CuO}=\dfrac{50.20}{100}=10\left(g\right)\)

\(m_{Fe_2O_3}=50-10=40\left(g\right)\)

\(n_{CuO}=\dfrac{m}{M}=\dfrac{10}{80}=0,125\left(mol\right)\)

\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{40}{160}=0,25\left(mol\right)\)

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

\(1mol\)     \(1mol\)

\(0,125mol\) \(0,125mol\)

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

\(1mol\)       \(3mol\)

\(0,25mol\)   \(0,75mol\)

\(V_{H_2}=n.22,4=\left(0,125+0,75\right).22,4=19,6\left(l\right)\)

nFe = 46,4/56 = 29/35 (mol)

PTHH: 4Fe + 3O2 -> (t°) 2Fe2O3

Mol: 29/35 ---> 87/140 ---> 29/70

mFe2O3 = 29/70 . 160 = 464/7 (g)

Vkk = 87/140 . 5 . 22,4 = 69,6 (l)

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

0,2----------0,6------0,4-----0,6 mol

n H2O=\(\dfrac{10,8}{18}\)=0,6 mol

=>VH2=0,6.22,4=13,44l

b)m Fe=0,4.56=22,4g

c) m Fe2O3=0,2.160=32g

\(n_{Fe}=\dfrac{2,8}{56}=0,05mol\)

\(\Rightarrow m_{Cu}=6-2,8=3,2g\)\(\Rightarrow n_{Cu}=0,05mol\)

\(CuO+H_2\rightarrow Cu+H_2O\) 

            0,05     0,05

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

               0,075    0,05

\(\Rightarrow\Sigma n_{H_2}=0,075+0,05=0,125mol\)

\(\Rightarrow V=0,125\cdot22,4=2,8l\)

CHÚC BẠN HỌC TỐT

a) Theo đề bài, ta có: \(n_{O2}=\dfrac{20}{32}=0,625\left(mol\right)\)

PTHH: \(2H_2+O_2\underrightarrow{o}2H_2O\)

pư............1.........0,5......1 (mol)

Ta có tỉ lệ: \(\dfrac{1}{2}< 0,625\). Vậy O2 dư, H2 hết.

\(\Rightarrow m_{H2O}=18.1=18\left(g\right)\)

Vậy.........

b) Đề nhầm bạn ơi, Cu chỉ có hóa trị là: I hoặc II