Excess Calculation-Solids & Solutions Excess

Introduction

• Excess calculations involve determining the quantity of a reactant in excess and the limiting reactant in a chemical reaction.
• In these calculations, we focus on the mass of a product formed when one reactant is in excess.

Worked Example

• Given: 2.45g of magnesium added to 100 cm³ of 1 mol/l hydrochloric acid (HCl).
• Reaction: Mg + 2HCl → MgCl₂ + H₂

Step 1: Calculate the Number of Moles

• Moles of Mg:
  • Moles = Mass / GFM (GFM of Mg = 24.5)
  • Moles = 2.45 / 24.5 = 0.1 moles
• Moles of HCl:
  • Moles = Concentration (C) x Volume (V) = 1 x 0.1 = 0.1 moles

Step 2: Determine the Quantity of Each Reactant Used

• Mg + 2HCl
  • 0.1 moles of Mg requires 0.2 moles of HCl.
  • Since there are only 0.1 moles of HCl, it's the limiting reactant.
  • Mg is in excess.

Step 3: Calculate the Mass of Hydrogen Produced

• Use the balanced equation:
  • Mg + 2HCl → MgCl₂ + H₂
  • 0.1 moles of Mg produces 0.1 moles of H₂.
• Calculate the mass of H₂:
  • Moles of H₂ = 0.1 moles
  • Mass = Moles x Molar mass (H₂ = 2g/mol)
  • Mass = 0.1 x 2 = 0.2g

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Digestion of proteins

Conclusion:

• In this reaction, when 2.45g of Mg reacts with 100 cm³ of 1 mol/l HCl, 0.2g of hydrogen gas is produced.