Question

12. 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 amount of oxygen, how many grams of each product are formed?

Could someone help me with this question thanks!

Answer 1

a) The balanced reaction is given below

b) 1 mole of chloroethylene reacts with 2.5 moles of oxygen

Molar mass of chloroethylene = 2 * 12 + 3 * 1 + 35.5 = 62.5 gm/mol

Number of moles of chloroethylene = 35/62.5 = 0.56 moles

Moles of O2 gas = 2.5 * 0.56 = 1.4 moles

c) Number of moles of chloroethylene = 25/62.5 = 0.4

Moles of CO2 formed = 2 * 0.4 = 0.8 moles

Moles of H2O formed = Moles of HCl formed = 0.4 moles

Mass of CO2 formed = 0.8 * 44 = 35.2 gms

Mass of H2O formed = 0.4 * 18 = 7.2 gms

Mass of HCl formed = 0.4 * 36.5 = 14.6 gms