Question

Answer 1

Câu 1:

\(S+O_2\underrightarrow{t^0}SO_2\\ 4P+O_2\underrightarrow{t^0}2P_2O_5\\ 2Zn+O_2\underrightarrow{t^0}2ZnO\\ CH_4+2O_2\underrightarrow{t^0}CO_2+2H_2O\)

Câu 3:

\(a,n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Pthh: 4Al + 3O₂ → (t^o) 2Al₂O₃

Mol:   0,2      0,15          0,1

\(b,m_{Al}=0,2.27=5,4\left(g\right)\\ c,V_{O_2}=0,15.22,4=3,36\left(l\right)\\ V_{kk}=3,36.5=16,8\left(l\right)\)

Answer 2

Câu 2 :

a. \(n_{Cu}=\dfrac{19,2}{64}=0,3\left(mol\right)\)

PTHH : 2Cu + O₂ -> 2CuO

              0,3      0,15              ( mol )

\(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH : O₂ + 4H₂ -> 2H₂O

             0,05  0,2

Xét tỉ lệ : \(\dfrac{0,15}{1}>\dfrac{0,2}{4}\) => O₂ dư , H₂ hết

\(m_{O_2\left(dư\right)}=\left(0,15-0,05\right).32=3,2\left(g\right)\)

 

Answer 3

Câu 3 :

a. \(n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH : 4Al + 3O₂ -> 2Al₂O₃

               0,2   0,15         0,1       ( mol )

b) \(m_{Al}=0,2.27=5,4\left(g\right)\)

c) \(V_{O_2}=0,15.22,4=3,36\left(l\right)\)

\(\Rightarrow V_{kk}=3,36.5=16,8\left(l\right)\)

Answer 4

Bài 4 :

a. PTHH : 2Zn + O₂ -> 2ZnO

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

\(n_{O_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

 b. Xét tỉ lệ \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) => Zn đủ , O₂ dư

\(m_{O_2\left(dư\right)}=\left(0,2-0,1\right).32=3,2\left(g\right)\)

c) \(m_{O_2}=0,2.32=6,4\left(g\right)\)

Theo ĐLBTKL

\(m_{Zn}+m_{O_2}=m_{ZnO}\\ \Rightarrow m_{ZnO}=6,4+13=19,4\left(g\right)\)