\(A:CuO+H_2\underrightarrow{t^0}Cu+H_2O\\ B.n_{Cu}=\dfrac{12,8}{64}=0,2mol\\ n_{CuO}=n_{H_2}=n_{H_2O}=0,2mol\\ m_{CuO}=0,2.80=16g\\ V_{H_2}=0,2.22,4=4,48l \\ m_{H_2O}=0,2.18=3,6g\)

a) \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)

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

             0,2--->0,2---->0,2

\(\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)

b) \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c) \(2H_2+O_2\xrightarrow[]{t^o}2H_2O\)

   0,2--->0,1

\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)

a: CuO+H2->Cu+H2O

0,2       0,2           0,2   0,2

mCu=0,2*64=12,8(g)

b: V=0,2*22,4=4,48(lít)

a, \(n_{H_2}=\dfrac{3,7185}{24,79}=0,15\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)

b, Theo PT: \(n_{FeSO_4}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{FeSO_4}=0,15.152=22,8\left(g\right)\)

c, \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,15}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow n_{CuO\left(dư\right)}=0,3-0,15=0,15\left(mol\right)\)

Chất rắn thu được sau pư gồm Cu và CuO dư.

⇒ m _{chất rắn} = m_Cu + m_{CuO (dư)} = 0,15.64 + 0,15.80 = 21,6 (g)

a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

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

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Cu}=0,1.64=6,4\left(g\right)\)

`CuO + H_2` $\xrightarrow[]{t^o}$ `Cu + H_2 O`

  `2`            `2`          `2`         `2`                `(mol)`

`n_[CuO] = 160 / 80 = 2 (mol)`

`a) V_[H_2] = 2 . 22,4 = 44,8 (l)`

`b) m_[Cu] = 2 . 64 = 128 (g)`

`c) m_[H_2 O] = 2 . 18 = 36 (g)`

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)

\(n_{Cu}=\dfrac{8}{80}=0,1mol\)

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

 0,02      0,06             0,04                  ( mol )

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

 0,1      0,1            0,1                ( mol )

\(m_{Fe}=0,04.56=2,24g\)

\(m_{Cu}=0,1.64=6,4g\)

\(n_{H_2}=0,06+0,1=0,16mol\)

\(n_{Fe_2O_3}=\dfrac{3,2}{160}=0,02mol\)

\(m_{CuO}=\dfrac{8}{80}=0,1mol\)

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

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

\(m_{Fe}=0,02\cdot2\cdot56=2,24g\)

\(m_{Cu}=0,1\cdot64=6,4g\)

\(\Sigma n_{H_2}=0,02\cdot3+0,1=0,16mol\Rightarrow V_{H_2}=3,584l\)

a) Fe tham gia phản ứng, $m_{Fe} = 6 - 3,2 =2,8(gam)$

b) $Fe + 2HCl \to FeCl_2 + H_2$

Theo PTHH : $n_{H_2} = n_{Fe} = \dfrac{2,8}{56} = 0,05(mol)$

$V_{H_2} = 0,05.24,79 = 1,2395(lít)$