liquid C2H3Cl CO2 H2O HCl gas 35.00 g 25.00 g

liquid C2H3Cl CO2 H2O HCl gas 35.00 g 25.00 g

 

 

The combustion of liquid chloroethylene, C2H3Cl,

yields carbon dioxide, steam, and hydrogen chloride gas.

a) Write a balanced equation for the reaction.

b) How many moles of oxygen are required

to react with 35.00 g of chloroethylene?

c) If 25.00 g of chloroethylene react with an excess of oxygen,

how many grams of each product are formed?

 

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a) Write a balanced equation for the reaction.

 

C2H3Cl + O2 → CO2 + H2O + HCl

 

 

1단계: 반응물 C2H3Cl의 2C → 생성물 2CO2

2단계: 생성물 2CO2 + H2O의 5O → 반응물 5/2 O2

 

C2H3Cl + 5/2 O2 → 2CO2 + H2O + HCl

 

 

3단계: 양번에 2씩 곱하면,

 

2C2H3Cl + 5O2 → 4CO2 + 2H2O + 2HCl

 

 

 

 

b) How many moles of oxygen are required

to react with 35.00 g of chloroethylene?

 

C2H3Cl의 몰질량 = 62.50 g/mol

 

 

C2H3Cl : O2 = 2 : 5 = (35.00/62.50 mol) : ? mol

? = 5 × (35.00/62.50) / 2 = 1.400 mol O2

 

 

 

 

c) If 25.00 g of chloroethylene react with an excess of oxygen,

how many grams of each product are formed?

 

C2H3Cl : CO2 = 2 : 4 계수비(= 몰수비) 이므로,

25.00/62.50 mol C2H3Cl 반응 시 생성되는 CO2의 몰수를 계산하면,

C2H3Cl : CO2 = 2 : 4 = 25.00/62.50 mol : ? mol

? = 4 × (25.00/62.50) / 2 = 0.8 mol CO2

 

CO2의 몰질량 = 44.01 g/mol 이므로,

0.8 mol CO2의 질량을 계산하면,

0.8 mol × (44.01 g/mol) = 35.21 g CO2

 

 

(25.00/62.50 mol) × (4/2) × (44.01 g/mol) = 35.21 g CO2

 

 

(25.00/62.50 mol) × (2/2) × (18.015 g/mol) = 7.206 g H2O

 

 

(25.00/62.50 mol) × (2/2) × (36.46 g/mol) = 14.58 g HCl