`PTHH:`

  `CuO + H_2 -t^o-> Cu + H_2 O`

`0,045`   `0,045`       `0,045`             `(mol)`

`n_[CuO]=[3,6]/80=0,045(mol)`

`a)m_[Cu]=0,045.64=2,88(g)`

`b)V_[H_2]=0,045.22,4=1,008(l)`

CuO+H2-to>Cu+H2O

0,03--0,03----0,03---0,03

n CuO=\(\dfrac{2,4}{80}\)=0,03 mol

=>VH2=0,03.22,4=0,672l

2Al+3H2SO4->Al2(SO4)3+3H2

0,02--------------------------------0,03

=>m Al=0,02.27=0,54g

\(n_{CuO}=\dfrac{2,4}{80}=0,03\left(mol\right)\\ PTHH:CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ n_{H_2}=n_{Cu}=n_{CuO}=0,03\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\\ b,m_{Cu}=0,03.64=1,92\left(g\right)\)

Bài 1:

a,n_CuO = 

Chúc em học tốt

a) PTHH: H₂ + CuO \(\underrightarrow{t^o}\) Cu + H₂O(1)
b) n_CuO = \(\frac{4,8}{80}=0,06\left(mol\right)\)
Theo PT(1): n_Cu = n_CuO = 0,06 (mol)
=> m_Cu = 0,06.64 = 3,84 (g)
c) Theo PT(1): n_{\(H_2\)} = n_Cu = 0,06 (mol) = n_{\(H_2\)(2)}
=> V_{\(H_2\)} = 0,06.22,4 =1,344 (l)
d) PTHH: Fe + 2HCl \(\rightarrow\) FeCl₂ + H₂\(\uparrow\)(2)
Theo PT(2): n_Fe = n_{\(H_2\)} = 0,06 (mol)
=> m_Fe = 0,06.56 = 3,36 (g)
Theo PT(2): n_HCl = 2n_{\(H_2\)} = 0,12 (mol)
=> m_HCl = 0,12.36,5 = 4,38(g)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{HCl}=1\cdot0,2=0,2\left(mol\right)\\ \Rightarrow n_{H_2}=\dfrac{1}{2}n_{HCl}=0,1\left(mol\right)\\ \Rightarrow V_{H_2\left(đktc\right)}=0,1\cdot22,4=2,24\left(l\right)\\ b,PTHH:CuO+H_2\rightarrow^{t^o}Cu+H_2O\\ \Rightarrow n_{Cu}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)

`a)`

PTHH : `Zn + 2HCl -> ZnCl_2 + H_2`

`200ml = 0,2l`

`n_{HCl} = 0,2 . 1 = 0,2` `mol`

`n_{H_2} = 1/2 . n_{HCl} = 0,1` `mol`

`V_{H_2} = 0,1 . 22,4 = 2,24` `l`

`b)`

`CuO + H_2 -> Cu + H_2O`

Ta có : `n_{H_2} = 0,1` `mol`

`-> n_{Cu} = n_{H_2} = 0,1` `mol`

`-> m_{Cu} = 0,1 . 64 = 6,4` `gam`

PTHH : \(CuO+H_2-t^o->Cu+H_2O\)

----------0,03-----0,03------------0,03----0,03 ( mol )

Ta có n_CuO= 2,4/80 =0,03

theo PTHH => n_{H2 =0,03}

=> V_{H2 =0,03.22,4=0,672 lít}

b) \(2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\)

n_H2 =0,03 mol.Theo PTHH => n_Al =0,02 mol

=> m_Al =0,02.27=0,54 gam

=

PTHH: CuO + H₂ -t^o-> Cu + H₂O (1)

Ta có: \(n_{CuO}=\dfrac{2,4}{80}=0,03\left(mol\right)\)

a, \(n_{H_2}=n_{CuO}=0,03\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,03.22,4=0,672\left(l\right)\)

b, PTHH: 2Al + 3H₂SO₄ -> Al₂(SO₄)₃ + 3H₂ (2)

Ta có: \(n_{H_2\left(2\right)}=n_{H_2\left(1\right)}=0,03\left(mol\right)\\ =>n_{Al\left(2\right)}=\dfrac{2.0,03}{3}=0,02\left(mol\right)\\ =>m_{Al}=0,02.27=0,54\left(g\right)\)

a) \(CuO+H_2-^{t^o}\rightarrow Cu+H_2O\)

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

\(\Rightarrow n_{CuO\left(pứ\right)}=0,2.80\%=0,16\left(mol\right)\)

\(n_{H_2O}=n_{CuO}=0,16\left(mol\right)\)

=> \(m_{H_2O}=0,16.18=2,88\left(g\right)\)

b) \(n_{H_2}=0,15\left(mol\right)\)

Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,15}{1}\Rightarrow\)Sau phản ứng CuO dư

Chất rắn sau phản ứng là Cu, CuO dư

\(m_{cr}=0,15.64+\left(0,2-0,15\right).80=13,6\left(g\right)\)

c) Gọi x là số mol CuO phản ứng 

\(m_{cr}=\left(0,2-x\right).80+64x=13,28\)

=> x=0,17 (mol)

\(H=\dfrac{0,17}{0,2}.100=85\%\)