Excess Reagent Calculation in Chemical Reaction

How to identify the excess reagent and calculate the amount that remains after a reaction?

Given 36 liters of hydrogen and 12 liters of carbon disulfide in a reaction producing methane and hydrogen sulfide, how do we determine the excess reagent and the remaining amount after the reaction?

Answer:

Carbon disulfide is the excess reagent in this reaction. After the reaction, 3 liters of carbon disulfide remains unreacted.

During a chemical reaction, it is important to identify the excess reagent to understand the reaction's efficiency and the amount of reactants consumed. In the given scenario, hydrogen and carbon disulfide react to produce methane and hydrogen sulfide. The balanced chemical equation is:

4H₂(g) + CS₂(g) -> CH₄(g) + 2H₂S(g)

Given the initial amounts of 36 liters of H₂ and 12 liters of CS₂, we need to determine which reagent is in excess and calculate the remaining amount of that reagent after the reaction.

To solve this problem, we first need to determine the stoichiometry of the reaction. According to the balanced equation, 4 moles of H₂ react with 1 mole of CS₂ to produce 1 mole of CH₄ and 2 moles of H₂S. Using the molar volume of an ideal gas at Standard Temperature and Pressure (STP), which is approximately 22.4 liters per mole, we can calculate the amount of each reagent required for a complete reaction.

For the given amounts of reactants:

- 36 liters of H₂ requires 9 liters of CS₂ for a complete reaction.

- Since there are 12 liters of CS₂ initially, it is the excess reagent in this reaction.

- After the reaction, 3 liters of CS₂ remains unreacted, indicating that it was not fully consumed in the reaction.

By understanding the concept of excess reagent and performing the necessary calculations, we can determine the efficiency of a chemical reaction and the amount of reactants that remain after the reaction is completed.

For further clarification and examples on excess reagent calculations, you can refer to additional resources on this topic.

← Understanding newton s second law of motion Structural formulas for various compounds →