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

a) 

\(KClO_3\xrightarrow[MnO_2]{t^o}KCl+\dfrac{3}{2}O_2\uparrow\)

\(\dfrac{1}{2}O_2+Cu\xrightarrow[]{t^o}CuO\)

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

\(H_2O+Na_2O\rightarrow2NaOH\)

b)

\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

\(H_2O\xrightarrow[]{đp}H_2+\dfrac{1}{2}O_2\)

\(FeO+H_2\xrightarrow[]{t^o}Fe+H_2O\)

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

a) (1) 2 KClO3 -to-> 2 KCl + 3 O2

(2) Cu +1/2 O2 -to-> CuO

(3) CuO + H2 -to-> Cu + H2O

(4) Na2O + H2O -> 2 NaOH

b) (1) Cu(OH)2 -to-> CuO + H2O

(2) Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O

(3) H2O -đp-> H2 + 1/2 O2

(4) 3 H2 + Fe2O3 -to-> 2 Fe +3 H2O

(5) Fe + H2SO4 -> FeSO4 + H2

Giup em với nha mn

a) (1) 2 KClO3 -to-> 2 KCl + 3 O2

(2) Cu +1/2 O2 -to-> CuO

(3) CuO + H2 -to-> Cu + H2O

(4) Na2O + H2O -> 2 NaOH

b) (1) Cu(OH)2 -to-> CuO + H2O

(2) Fe3O4 + 4 H2 -to-> 3 Fe + 4 H2O

(3) H2O -đp-> H2 + 1/2 O2

(4) 3 H2 + Fe2O3 -to-> 2 Fe +3 H2O

(5) Fe + H2SO4 -> FeSO4 + H2

\(a.2Al+3CuO-^{t^o}>Al_2O_3+3Cu\)

\(b.2Na+2H_2O->2NaOH+H_2\)

\(c.Cu+2H_2SO_{4_{dn}}->CuSO_4+SO_2+2H_2O\)

CuO+H2--->Cu+H2O

Tỉ lệ: 1:1:1:1

CuO+H₂->Cu+H₂O(cái này cân =luôn r bn)

Tỉ lệ:

Số phân tử CuO:Số phân tử H₂:Số nguyên tử Cu:Số phân tử H₂O=1:1:1:1

PTHH:

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

tỉ lệ: CuO : H₂ : Cu: H₂O = 1:1:1:1

chúc bn học tốt

\(a,4K+O_2\underrightarrow{t^o}2K_2O\\ 2Cu+O_2\underrightarrow{t^o}2CuO\\ b,CuO+H_2\underrightarrow{t^o}Cu+H_2O\\ Ag_2O+H_2\underrightarrow{t^o}2Ag+H_2O\\ c,2K+2H_2O\rightarrow2KOH+H_2\\ Na_2O+H_2O\rightarrow2NaOH\\ BaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(n_{H_2}=\dfrac{5.6}{22.4}=0.25\left(mol\right)\)

\(n_{CuO}=\dfrac{7.2}{80}=0.09\left(mol\right)\)

\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)

Lập tỉ lệ : 

\(\dfrac{0.25}{1}>\dfrac{0.09}{1}\rightarrow H_2dư\)

\(n_{Cu}=n_{CuO}=0.09\left(mol\right)\)

\(m=0.09\cdot64=5.76\left(g\right)\)

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ n_{CuO}=\dfrac{7,2}{80}=0,09\left(mol\right)\\ PTHH:CuO+H_2\underrightarrow{to}Cu+H_2O\\ Vì:\dfrac{0,25}{1}>\dfrac{0,09}{1}\\ \rightarrow H_2dư\\ \rightarrow n_{Cu}=n_{CuO}=0,09\left(mol\right)\\ m_{Cu}=0,09.64=5,76\left(g\right)\)