Particle Diagram

Hydrogen and Oxygen are both diatomic.

Often times, Hydrogen is drawn an empty circle but Oxygen tends to be a red circle.

Particle Diagram

A molecule of water tends to be drawn with two circles and one red circle.

Finding the Limiting Reactant

According to the problem, before they began there were 4 molecules of both hydrogen gas and oxygen gas.

Finding the Limiting Reactant

Finding the Limiting Reactant

​Particle Diagram: 

According to the diagram, 4 water molecules were made, all of the hydrogen gas was consumed, and there were 2 molecules of oxygen leftover.

Finding the Limiting Reactant

You will need a piece of paper to take notes.

Watch the clip.

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​Let's step it up a notch.

20.0g of CaCl₂

20.0g of AgNO₃

Converting to Moles

Balanced equation allow us to use mole ratios, but before that, all mass needs to be converted into moles.

Finding the Limiting Reactant

The ratio of AgNO₃ to CaCl₂ is 2:1. Therefore, if .180 moles of CaCl₂ is used, that would mean 2x more AgNO₃ is needed (.360 moles). Is there enough?

No! So CaCl₂ is not the limiting reactant.

Finding the Limiting Reactant

The ratio of AgNO₃ to CaCl₂ is 2:1. Therefore, if .118 moles of AgNO₃ is used, that would mean 1/2x CaCl₂ is needed (.059 moles). Is there enough?

Yes! So AgNO₃ is the limiting reactant.

Finding the Limiting Reactant

The ratio of AgNO₃ to AgCl is 1:1. Therefore, if .118 moles of AgNO₃ is used, that would mean the same amount of AgCl is made.

Finding the Limiting Reactant

The ratio of AgNO₃ to Ca(NO₃)₂ is 2:1. Therefore, if .118 moles of AgNO₃ is used, that would mean 1/2x Ca(NO₃)₂ is made (.059 moles).

Let's write down the BCA table

B: .118 .180 0 0
C: -.118 -.059 +.059 +.180
A: 0 .121 .059 .180

Now convert the moles into grams using the molar mass. You can google them. :)

On your own.

Problem #2, Page 6