Bài 1 : 

a)

\(2Cu + O_2 \xrightarrow{t^o} 2CuO\)

b)

Ta có :

\(n_{Cu} = \dfrac{32}{64} = 0,5(mol)\)

Theo PTHH : \(n_{O_2} = 0,5n_{Cu} = 0,25(mol)\\ \Rightarrow V_{O_2} = 0,25.22,4 = 5,6(lít)\)

c) Ta có : \(n_{CuO} = n_{Cu} = 0,5(mol)\Rightarrow m_{CuO} = 0,5.80 = 40(gam)\)

Bài 2 : 

 \(Zn + H_2SO_4 \to ZnSO_4 + H_2\)

Theo PTHH : 

\(n_{H_2SO_4} = n_{H_2} = n_{Zn} =\dfrac{13}{65} = 0,2(mol)\)

Suy ra :

\(V_{H_2} = 0,2.22,4 = 4,48(lít)\\ m_{H_2SO_4} = 0,2.98 = 19,6(gam)\)

Bài 1 :

a) PTHH : \(2Cu+O_2-t^o->2CuO\)

b) \(n_{Cu}=\frac{32}{64}=0,5\left(mol\right)\)

Theo pthh : \(n_{O2}=\frac{1}{2}n_{Cu}=0,25\left(mol\right)\)

=> \(V_{O2}=0,25\cdot22,4=5,6\left(l\right)\)

c) Theo pthh : \(n_{CuO}=n_{Cu}=0,5\left(mol\right)\)

=> \(m_{CuO}=80\cdot0,5=40\left(g\right)\)

Bài 2 : 

a) \(PTHH:Zn+H_2SO_4-->ZnSO_4+H_2\uparrow\)

b) \(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)

Theo pthh : \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\)

=> \(V_{H2}=0,2\cdot22,4=4,48\left(l\right)\)

c) Theo pthh : \(n_{H2SO4\left(pứ\right)}=n_{Zn}=0,2\left(mol\right)\)

=> \(m_{H2SO4\left(pứ\right)}=98\cdot0,2=19,6\left(g\right)\)

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

b)\(n_{CuO}=\dfrac{4}{80}=0,05\left(m\right)\)

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

tỉ lệ          :1       1           1           1

số mol     :0,05  0,05      0,05     0,05           

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

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

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{H_2SO_4}= n_{H_2}= n_{Zn} = \dfrac{19,5}{65} =0,3(mol)\\ m_{H_2SO_4} = 0,3.98 = 29,4(gam)\\ c) V_{H_2} = 0,3.22,4 = 6,72(lít)\\ d) Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ n_{Fe} = \dfrac{2}{3}n_{H_2} = 0,2(mol)\\ m_{Fe} = 0,2.56 = 11,2(gam)\)

\(a) Zn + H_2SO_4 \to ZnSO_4 + H_2\\ b) n_{Zn} = \dfrac{13}{65} = 0,2 < n_{H_2SO_4} = \dfrac{49}{98} = 0,5 \to H_2SO_4\ dư\\ c) n_{H_2}= n_{Zn} = 0,2(mol)\\ V_{H_2} = 0,2.22,4= 4,48(lít)\\ d) CuO + H_2 \xrightarrow{t^o} Cu + H_2O\\ n_{Cu} = n_{H_2}= 0,2(mol)\\ m_{Cu} = 0,2.64= 12,8(gam)\)

     a)PTHH(1): Zn+H2SO4 →ZnSO4+H2

     b)  theo gt: 0,21 < 0,51⇒H2SO4dư 0,21 < 0,51⇒H2SO4dư , tính số mol các chất theo Zn

ta có theo PTHH: nH2SO4= nZn=0,2(mol)

⇒ nH2SO4dư=0,5−0,2=0,3(mol)

⇒ mH2SO4dư=0,3⋅98=29,4(g)

 Do tỉ lệ phương trình là 1:1:1:1

=>  nH2 = nZn = 0,2 mol

=>  VH2= n.22,4= 0,2.22,4= 4,48 (l)

c) PTHH(2) : Cuo + H2  Cu + H2O

Theo PTHH (1)  và tích chất bắc cầu ta được

=>  nH2 = nCu = 0,2 mol

=>  mCu = n.M= 0,2.64= 12,8 (g)

a, Ta có:

n_Zn = 13/65= 0,2(mol)

PTHH: Zn + H₂SO₄ → ZnSO₄ + H₂

               _{0,2-----------------------------------0,2}

Theo PT : n_ZnSO4 = 0,2.1/1 = 0,2(mol)

m_ZnSO4 = 0,2. 161 = 32,2(g)

b, Ta có:

Theo PT : n_H2 = 0,2.1/1 = 0,2(mol)

V_H2(đktc) = 0,2 . 22,4 = 4,48(l)

CuO+H2-to>Cu+H2O

           0,2-----0,2

=>m Cu=0,2.64=12,8g

a) \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

\(n_{HCl}=0,2.1=0,2\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

Xét tỉ lệ:  \(\dfrac{0,2}{1}>\dfrac{0,2}{2}\) => Zn dư, HCl hết

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

__________0,2-------------->0,1

=> V_H2 = 0,1.22,4 = 2,24(l)

b)

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

______0,1->0,05

=> m_O2 = 0,05.22,4 = 1,12 (l)

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

a. \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

Theo PTHH: \(n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow m_{muối}=0,1.161=16,1\left(g\right)\)

b. \(n_{H_2thu.được}=n_{Zn}=0,1\left(mol\right)\)

\(H_2+\dfrac{1}{2}O_2\underrightarrow{t^o}H_2O\)

0,1    0,05

\(V_{O_2}=0,05.22,4=1,12\left(l\right)\)

\(\Rightarrow V_{không.khí}=1,12.5=5,6\left(l\right)\)

\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)

Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư

\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)