- Giống nhau :

+ Đều là các chất khí không màu, không mùi ở nhiệt độ phòng.

+ Đều có chứa liên kết đôi, phân tử gồm 2 nguyên tử.

- Khác nhau : 

a) m HCl \(0,5.\left(36,5\right)=18,25g\)

b) Zn +2HCl -> ZnCl2 +H2

  0,25...0,5...................0,25

V H2 = 0,25.22,4 =5,6 lít 

m Zn = 0,25.65=16,25g 

a) Có: \(n_{HCl}=0,5mol\)

\(\Rightarrow m_{HCl}=0,5.\left(1+35,5\right)=18,25g\)

b)

\(PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

Có: \(n_{HCl}=0,5mol\)

Theo PTHH:

\(\Rightarrow n_{H_2}=0,25mol\)

\(\Rightarrow V_{H_2}=0,25.22,4=5,6l\)

     \(n_{Zn}=0,25mol\)

\(\Rightarrow m_{Zn}=0,25.65=16,25g\)

CTHH: CuO

TL:

7.A

8.C

9.A

10.B

HT

k mình nha 

hahaha.....      7b 8c 9a 10a

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1. Bảng hóa trị một số nguyên tố hóa học.

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

Phương trình hóa học phản ứng

Zn + H₂SO₄ ---> ZnSO₄ + H₂

1   :      1       :       1         :  1

0.2                     0,2          0,2

mol                    mol          mol

=> \(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

b) \(m_{ZnSO_4}=n.M=0,2.161=32,2\left(g\right)\)

c) Ta có \(C\%=\frac{m_{ct}}{m_{dd}}.100\%=24,5\%\)

=> \(m_{ct}=\frac{C\%.m_{dd}}{100\%}=\frac{24,5\%.200}{100\%}=49\left(g\right)=m_{H_2SO_4}\)

=> \(m_{H_2O}=151\left(g\right)\)

=> \(n_{H_2SO_4}=\frac{m}{M}=\frac{49}{98}=0,5\)(mol) 

Dễ thấy \(\frac{n_{Zn}}{1}< \frac{n_{H_2SO_4}}{1}\)

=> H₂SO₄ dư 0,5 - 0,2 = 0,3 (mol) 

=> \(m_{H_2SO_4\text{ dư }}=n.M=0,3.98=29,4\left(g\right)\); \(m_{H_2SO4\text{ tham gia}}=n.M=0,2.98=19,6\)(g)

Áp dụng đinhk luật bảo toàn khối lượng 

=> \(m_{H_2SO_4}+m_{Zn}=m_{ZnSO4}+m_{H_2}\)

=> \(m_{H_2}=m_{H_2SO_4}+m_{Zn}-m_{ZnSO_4}=19,6+13-32,2=0,4\left(g\right)\)

=> \(m_{saupư}=m_{ZnSO_4}+m_{H_2SO_4\text{ dư}}+m_{H_2O}-m_{H_2}=32,2+29,4+151-0,4=232,2\left(g\right)\)

=> \(C\%_{H_2SO_4}=\frac{m_{ct}}{m_{sau\text{ pư}}}.100\%=\frac{29,4}{232,2}.100\%=12,66\%\)

\(C\%_{ZnSO_4}=\frac{m_{ct}}{m_{dd}}.100\%=\frac{32,2}{232,2}.100\%=13,87\%\)

2                                                                                                             

nha nha 

\(nC_2H_4=\frac{3,36}{22,4}=0,15mol\)

\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)

\(nO_2=3nC_2H_4=0,45mol\)

\(\rightarrow VO_2=0,45.24,79=11,1555l\)