Question: In the balanced equation Cu + 2 HCl --> CuCl2 + H2, if you start with 0.15mol Cu and 0.38mol HCl, how many moles of HCl are left over (which means unused)?

Answer: After performing the necessary calculations, the number of moles of HCl that are left over in the reaction between 0.15 mol of copper and 0.38 mol of HCl is 0.08 mol.

Understanding Stoichiometry in Chemistry: Calculating Leftover Moles in a Chemical Reaction

20 Nov 2023

Explanation:

Uses of Stoichiometry

In chemistry, stoichiometry is a crucial tool used to calculate relationships between moles and masses or moles of different substances involved in a chemical reaction. In this scenario, the reaction between 0.15 mol of copper and 0.38 mol of HCl can be analyzed using stoichiometry to determine the amount of HCl left over.

Calculating Moles of HCl Utilized

By applying the mole ratio of 1:2 between copper and HCl in the balanced equation, the moles of HCl utilized by 0.15 mol of copper can be determined:

0.15 mol Cu * (2 mol HCl / 1 mol Cu) = 0.30 mol HCl

This indicates that 0.30 mol of HCl is consumed in the reaction. Since 0.38 mol of HCl was initially present, the leftover or unused moles of HCl can be calculated:

Leftover = 0.38 mol - 0.30 mol = 0.08 mol

Final Calculation and Result

By subtracting the stoichiometrically required 0.30 mol of HCl for 0.15 mol of Cu from the initial 0.38 mol HCl, the remaining moles of HCl are found to be 0.08 mol. This unutilized amount represents the moles of HCl left over in the chemical reaction.

Therefore, 0.08 moles of HCl remains unused in the reaction.

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