PYTR Class

Learning Objectives

1. The use of mole concept to solve stoichiometric problems
2. Determination of limiting reactant

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Part 1: Limiting Reactant

In every reaction involving two reactants, one may become limiting and the other will be in excess. This means that…

• Limiting reactant has lower number of moles
• Limiting reactant limits the amount of product(s) produced
• Excess reactant will not have completely being used at the end of the reaction
• Some excess reactant will remain unreacted

Based on this graph, oxygen is the limiting reactant as it is completely used to form water. There are some amount of hydrogen gas remained unreacted at the end of this reaction.

Part 2: Calculating Limiting Reactant

Consider this reaction:

10 g of H₂ and 20 g of O₂

Which is the limiting reactant?

Step 1: Find n of each reactant based on the starting amount

Step 2: Use mole ratio

Reactant	O2	H2
Moles, n	0.625 moles	4.96 moles
Mole ratio (balanced equation)	1	2
Reaction ratio	0.625 / 1 = 0.625	4.96/2 =2.48

Based on this ratio, 0.626 (O₂) : 2.48 (H₂)

Oxygen will be the limiting reactant (having lower ratio)

You could also look at it using mole ratio as follow:

2 moles of H₂ react with 1 mole of O₂

For 0.625 moles of O₂​, we need:

Since we have 4.96 moles of H₂, which is more than 1.25 moles, O₂​ is the limiting reactant.

Part 3: Exercises

Question:

Find the amount of products formed when 5g of Mg and 50ml of 0.25 mol/dm³ of HCl reacted. The find the amount of excess or unreacted reactant.

Step 1: Write the Balanced Chemical Equation

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Step 2: Calculate Moles of Reactants

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Step 3: Identify the Limiting Reactant

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Step 4: Determine the Amount of Products Formed and Unreacted Reactant

Since HCl is limiting, we base our calculations on 0.0125 moles of HCl.

(a) Moles of H₂ gas produced:

(b) Moles of MgCl₂ produced:

Since HCl is the limiting reactant, there will be some excess Mg left unreacted: