Respuesta :
Final Answer:
The balanced chemical equation for the combustion of 1 mol of ethene (C₂H₄) with 4 mol of oxygen (O₂) is given by:
[tex]\[C₂H₄(g) + 3O₂(g) \rightarrow 2CO₂(g) + 2H₂O(v)\][/tex]
According to the stoichiometry of the reaction, 1 mol of ethene reacts with 3 mol of oxygen. In this scenario, 4 mol of oxygen are provided, which is in excess. Therefore, the limiting reactant is ethene, and the reaction will consume all available moles of ethene. The products formed are 2 mol of carbon dioxide (CO₂) and 2 mol of water (H₂O).
Hence, the number of moles of gases and vapors at the end of the reaction is 2 mol + 2 mol = 4 mol.
Therefore, the correct answer is (c) 4 mol.
Explanation:
In the given chemical equation, the stoichiometric coefficients represent the molar ratios between the reactants and products. For every 1 mol of ethene, 3 mol of oxygen are required. In the provided scenario, 4 mol of oxygen are in excess, meaning all ethene will be consumed in the reaction.
The balanced equation also indicates that for every mole of ethene, 2 moles of CO₂ and 2 moles of H₂O are produced. Therefore, with 1 mol of ethene, 2 mol of CO₂ and 2 mol of H₂O are formed, resulting in a total of 4 mol of gases and vapors.
In summary, the balanced equation and stoichiometry allow us to determine the quantities of reactants and products involved in the chemical reaction, leading to the conclusion that 4 mol of gases and vapors are produced.
