Question

Write the balanced chemical equation for the complete, stoichiometric combustion of ethylene in (a) nitrous oxide and (b) air. Compare the required number of moles and the oxidizer mass using each of the two oxidizers for the complete, stoichiometric combustion of ethylene.

Answer 1

Answer: Mass of nitrous oxide used is 264 grams and mass of oxygen gas used is 96 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$   ....(1)

• For a: Combustion of ethylene in nitrous oxide

The chemical equation for the combustion of ethylene in nitrous oxide follows the equation:

$C_2H_4+N_2O\rightarrow 2CO_2+2H_2O+6N_2$

By stoichiometry of the reaction;

6 moles of nitrous oxide are required for the complete combustion of ethylene molecule.

Calculating the mass of nitrous oxide using equation 1:

Molar mass of nitrous oxide = 44 g/mol

Moles of nitrous oxide = 6 moles

Putting values in equation 1, we get:

$6mol=\frac{\text{Mass of nitrous oxide}}{44g/mol}\\\\\text{Mass of nitrous oxide}=264g$

Thus, 264 grams of nitrous oxide are required for the complete combustion of ethylene.

• For b: Combustion of ethylene in air

The chemical equation for the combustion of ethylene in air follows the equation:

$C_2H_4+3O_2\rightarrow 2CO_2+2H_2O$

By stoichiometry of the reaction;

3 moles of oxygen are required for the complete combustion of ethylene molecule.

Calculating the mass of oxygen using equation 1:

Molar mass of oxygen = 32 g/mol

Moles of oxygen = 3 moles

Putting values in equation 1, we get:

$3mol=\frac{\text{Mass of oxygen}}{32g/mol}\\\\\text{Mass of oxygen}=96g$

Thus, 96 grams of oxygen are required for the complete combustion of ethylene.