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

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

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

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

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 

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

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

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

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

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

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

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

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

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

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

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

\(1mol\)     \(1mol\)

\(0,125mol\) \(0,125mol\)

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

\(1mol\)       \(3mol\)

\(0,25mol\)   \(0,75mol\)

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

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

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

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

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

`CuO+ H_2 -> Cu+ H_2O`

`0,03 ----0,03` mol

`Fe_2O_3+ 3H_2 ->2Fe + 3H_2O`

`0,1-------0,3` mol

`n_(CuO) = 2,4/80 =0,03` mol

`n_(Fe_2O_3)=16/160 =0,1` mol

`=> V_(H_2)=(0,3+0,03).22,4=7,392 l`

mFe2O3=80%.50=40(g)

=>nFe2O3=40/160=0,25(mol)

mCuO=50-40=10(g)

=>nCuO=10/80=0,125(mol)

Fe2O3+3H2--t*-->2Fe+3H2O

0,75____0,75

CuO+H2--t*-->Cu+H2O

0,125__0,125

ΣH2=ΣH2=0,75+0,125=0,875(mol)

=>VH2=0,875.22,4=19,6(l)

từ đó bạn suy ra =>

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

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

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

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

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 

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