Câu 7

\(a.Na+H_2O\xrightarrow[]{}NaOH+H_2\\ K_2O+H_2O\xrightarrow[]{}2KOH\\ SO_2+H_2O\xrightarrow[]{}H_2SO_3\\ P_2O_5+3H_2O\xrightarrow[]{}2H_3PO_4\)

\(b.Fe_2O_3+3H_2\xrightarrow[t^0]{}2Fe+3H_2O\\ HgO+H_2\xrightarrow[]{t^0}Hg+H_2O\\ CuO+H_2\xrightarrow[]{t^0}Cu+H_2O\\ PbO+H_2\xrightarrow[]{t^0}Pb+H_2O\)

Câu 8

\(\left(1\right)2KClO_3\xrightarrow[]{t^0}2KCl+3O_2\)(tác dụng nhiệt, xúc tác)

\(\left(2\right)Al+H_2SO_4\xrightarrow[]{}Al_2\left(SO_4\right)_3+H_2\)(nhiệt độ phòng)

-(1)Phản ứng phân huỷ.

-(2)Phản ứng thế.

-(1)Điều chế khí \(O_2\) trong phòng thí nghiệm.

-(2)Điều chế \(H_2\) trong phòng thí nghiệm

a) Fe + H₂SO₄ --> FeSO₄ + H₂

b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ --> FeSO₄ + H₂

           0,1------------------------>0,1

=> V_H2 = 0,1.22,4 = 2,24 (l)

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

PTHH: CuO + H₂ --t^o--> Cu + H₂O

Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\) => CuO dư, H₂ hết

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1<--0,1------->0,1

=> m_CuO(dư) = (0,15 - 0,1).80 = 4 (g)

m_Cu = 0,1.64 = 6,4 (g)

a, PT: \(Fe+H_2SO_4\rightarrow H_2SO_4+H_2\)

b, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

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

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: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,1\left(mol\right)\)

\(\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\Rightarrow m_{CuO\left(dư\right)}=0,05.80=4\left(g\right)\)

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

Bạn tham khảo nhé!

Bài 1.

\(n_{Fe}=\dfrac{5,6}{56}=0,1mol\)

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

0,1         0,1         0,1          0,1

\(V_{H_2}=0,1\cdot22,4=2,24l\)

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

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

0,15      0,1

\(\Rightarrow CuO\) dư và dư \(\left(0,15-0,1\right)\cdot80=4g\)

Bài 2.

\(n_P=\dfrac{3,1}{31}=0,1mol\)

\(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

0,1     0,125

\(V_{O_2}=0,125\cdot22,4=2,8l\)

\(n_{Zn}=\dfrac{19,5}{65}=0,3mol\)

\(2Zn+O_2\underrightarrow{t^o}2ZnO\)

0,3       0,125  0

0,25     0,125  0,25

0,05       0       0,25

\(\Rightarrow ZnO\) dư và dư \(0,05\cdot81=4,05g\)

Bài 1.

a, \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: Fe + H₂SO₄ ---> FeSO₄ + H₂

Mol:     0,1                                    0,1

b, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

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

Ta có: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\) ⇒ CuO dư, H₂ hết

PTHH: CuO + H₂ ---t^o----> Cu + H₂O

Mol:      0,1     0,1

\(m_{CuOdư}=\left(0,15-0,1\right).80=4\left(g\right)\)

CuO + H2 -to-> Cu + H2O

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

Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O

PbO + H2 -to-> Pb + H2O

HgO + H2 -to-> Hg + H2O

FeO + H2 -to-> Fe + H2O

a) 

FeO + H₂ --t^o--> Fe + H₂O

CuO + H₂ --t^o--> Cu + H₂O

b) \(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

PTHH: CuO + H₂ --t^o--> Cu + H₂O

              0,1<--0,1<-----0,1

=> \(m_{FeO}=12-0,1.80=4\left(g\right)\)

=> \(n_{FeO}=\dfrac{4}{72}=\dfrac{1}{18}\left(mol\right)\)

 FeO + H₂ --t^o--> Fe + H₂O

 \(\dfrac{1}{18}\)-->\(\dfrac{1}{18}\)----->\(\dfrac{1}{18}\)

=> \(V_{H_2}=\left(0,1+\dfrac{1}{18}\right).22,4=\dfrac{784}{225}\left(l\right)\)

c) \(m_{Fe}=\dfrac{1}{18}.56=\dfrac{28}{9}\left(g\right)\)

d) \(\left\{{}\begin{matrix}m_{CuO}=0,1.80=8\left(g\right)\\m_{FeO}=4\left(g\right)\end{matrix}\right.\)

b) ... :) ?

\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)

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

0,1                           0,1            ( mol )

\(\left\{{}\begin{matrix}m_{CuO}=0,1.80=8g\\m_{FeO}=12-8=4g\end{matrix}\right.\)

PTHH: H2+FeO-----to-----> Fe+H2O

            H2+CuO------to---->Cu+H2O

$a)Zn + 2HCl \to ZnCl_2 + H_2$
b)

Theo PTHH :

n H2 = n Zn = 1,3/65 = 0,02(mol)
V H2 = 0,02.22,4 = 0,448(lít)

c) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

Ta thấy :

n CuO = 2,4/80 = 0,03 > n H2 = 0,02 nên CuO dư

Theo PTHH :

n CuO pư = n Cu = n H2 = 0,02(mol)
Sau phản ứng có :

m Cu = 0,02.64 = 1,28(gam)

m CuO dư = 2,4 - 0,02.80 = 0,8(gam)

Ta có: \(n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

_____0,02_________________0,02 (mol)

b, V_H2 = 0,02.22,4 = 0,448 (l)

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

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

Xét tỉ lệ: \(\dfrac{0,03}{1}>\dfrac{0,02}{1}\), ta được CuO dư.

Theo PT: \(n_{Cu}=n_{CuO\left(pư\right)}=n_{H_2}=0,02\left(mol\right)\)

⇒ n_{CuO (dư)} = 0,01 (mol)

\(\Rightarrow m_{CuO\left(dư\right)}=0,01.80=0,89\left(g\right)\)

\(m_{Cu}=0,02.64=1,28\left(g\right)\)

Bạn tham khảo nhé!

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

b+c) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)=n_{CuO}=n_{Cu}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Cu}=0,1\cdot64=6,4\left(g\right)\\m_{CuO}=80\cdot0,1=8\left(g\right)\end{matrix}\right.\)

d) Ta có: \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)

\(\Rightarrow\) CuO còn dư, Hidro p/ứ hết

\(\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\) \(\Rightarrow m_{CuO\left(dư\right)}=80\cdot0,05=4\left(g\right)\)