Giải :
Dẫn khí hidro đi qua 20,25g kẽm oxit(ZnO), đun nóng vừa đủ để thu được kẽm kim loại và nước. Thể tích khí hidro cần dùng ở đktc là:

A.2,24l   B.11,2l   C.5,6l  D.0,56l

Học tốt !!!!!!!!!!!

A.2,24l

ko biet

Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

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

\(PTHH:\)

\(Fe+2HCl--->FeCl_2+H_2\left(1\right)\)

\(CuO+H_2\overset{t^o}{--->}Cu+H_2O\left(2\right)\)

a. Theo PT₍₁₎: \(n_{H_2}=n_{Fe}=0,15\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,15.22,4=3,36\left(lít\right)\)

b. Ta thấy: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\)

Vậy CuO dư.

Theo PT₍₂₎: \(n_{Cu}=n_{H_2}=0,15\left(mol\right)\)

\(\Rightarrow m_{Cu}=0,15.64=9,6\left(g\right)\)

Mik làm rồi nhé

Bn vuối xuống dưới là thấy

\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
pthh: \(H_2+CuO\underrightarrow{t^o}Cu+H_2O\)
        0,1                 0,1       0,1
=>\(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\m_{H_2O}=0,1.18=1,8\left(g\right)\end{matrix}\right.\)

PTHH: \(ZnO+H_2-t^o->Zn+H_2O\)

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

Ta có n_ZnO= 8,1/81 = 0,1 mol

Theo Ptth => n_H2 =0,1 mol

=> V_H2 =0,1 .22,4 = 2,24 lít

Theo PTHH ta có n_Zn = 0,1 mol

=> m_Zn =0,1.65=6,5 gam

\(Zn+2HCl->ZnCl_2+H_2\)

m_HCl =200.7,3%= 14,6 gam

ta có n_HCl = 14,6/36,5=0,4 mol

Ta có \(\dfrac{0,1}{1}< \dfrac{0,4}{2}=>HCl,dư\)

Theo PTHH ta có n_H2 = n_Zn =0,1 mol

=> V_H2 =0,1 .22,4 =2,24 lít

Vậy...

Ta có: \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

PTHH: ZnO + H₂ -t^o-> Zn + H₂O (1)

a, Ta có: \(n_{H_2}=n_{Zn}=n_{ZnO}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

b, \(m_{Zn}=0,1.65=6,5\left(g\right)\)

c, PTHH: Zn + 2HCl -> ZnCl₂ + H₂ (2)

Theo các PTHH và đề bài, ta có:

\(n_{Zn\left(2\right)}=n_{Zn\left(1\right)}=0,1\left(mol\right)\\ m_{HCl}=\dfrac{7,3.200}{100}=14,6\left(g\right)\\ =>n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

Ta có: \(\dfrac{n_{Zn\left(đề\right)}}{n_{Zn\left(PTHH\right)}}=\dfrac{0,1}{1}< \dfrac{n_{HCl\left(đề\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,4}{2}\)

=> Zn hết, HCl dư.

=> \(n_{H_2\left(2\right)}=n_{Zn\left(2\right)}=0,1\left(mol\right)\\ =>V_{H_2\left(2,đktc\right)}=0,1.22,4=2,24\left(l\right)\)

PTHH: ZnO + H₂ -t^o-> Zn + H₂O (1)

\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

a, \(n_{H_2}=n_{Zn}=n_{ZnO}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ b,m_{Zn}=0,1.65=6,5\left(g\right)\\ c,PTHH:Zn+2HCl->ZnCl_2+H_2\left(2\right)\)

Theo PTHH và đề bài, ta có: \(m_{HCl}=\dfrac{7,3.200}{100}=14,6\left(g\right)\\ =>n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\\ n_{Zn\left(2\right)}=n_{Zn\left(1\right)}=0,1\left(mol\right)\\ =>\dfrac{n_{HCl\left(đề\right)}}{n_{HCl\left(PTHH\right)}}=\dfrac{0,4}{2}>\dfrac{n_{Zn\left(đề\right)}}{n_{Zn\left(PTHH\right)}}=\dfrac{0,1}{1}\)

=> HCl dư, Zn hết, tính theo n_Zn

=> \(n_{H_2}=n_{Zn\left(2\right)}=0,1\left(mol\right)\\ =>V_{H_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\)

\(a) ZnO + H_2 \xrightarrow{t^o} Zn + H_2O\\ n_{Zn} = n_{H_2} = n_{ZnO} = \dfrac{16,2}{81} = 0,2(mol)\\ m_{Zn} = 0,2.65 = 13(gam)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ b) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2} = n_{CuO} = \dfrac{4}{80} = 0,05(mol)\\ m_{Cu} = 0,05.64 = 3,2(gam)\\ V_{H_2} = 0,05.22,4 = 1,12(lít)\)

a.b.

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

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,05     0,1                        0,05    ( mol )

\(V_{H_2}=0,05.22,4=1,12l\)

\(m_{HCl}=0,1.36,5=3,65g\)

c.

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

             0,05           0,05           ( mol )

\(m_{Cu}=0,05.64=3,2g\)

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