Question

Equimolar amounts of Cl₂ (g) and O₂ (g) are injected into a sealed rigid container, where they react according to the equation below:

2Cl₂ (g) + O₂ (g) ⇄ 2ClO (g) + 161 kJ/mol

If 6.4 moles of O₂ (g) undergo the reaction with excess Cl₂ (g), how many moles of ClO (g) are produced?

Answer 1

To calculate the number of moles of ClO produced from 6.4 moles of O₂ reacting with excess Cl₂, one must refer to the balanced chemical equation and apply stoichiometry. Assuming a 1:1 reaction, 6.4 moles of O₂ would theoretically yield 6.4 moles of ClO gas.

Step-by-step explanation:

The student's question pertains to the stoichiometry of a chemical reaction involving chlorine gas (Cl₂) and oxygen gas (O₂), and how to calculate the amount of ClO (g) produced given equimolar amounts of the reactants. Assuming 6.4 moles of O₂ reacting with an excess of Cl₂, we must refer to the balanced chemical equation for the reaction to determine the mole ratio of O₂ to ClO produced. However, the reaction for the formation of ClO is not provided in the question or reference material.

If we assume a hypothetical balanced equation, for instance, Cl₂ (g) + O₂ (g) → 2 ClO (g), we can then say in a 1:1 reaction ratio, 6.4 moles of O₂ would produce 6.4 moles of ClO. Since the Cl₂ is in excess, we don't need to calculate its moles. We utilize conversion factors and the molar stoichiometry to find the amount of product formed from a given amount of reactant. In summary, the number of moles of ClO produced would be equal to the number of moles of O₂ used if the reaction between Cl₂ and O₂ produces ClO in a 1:1 ratio.