Limiting Reactants

Let's start on Page 5 of the packet.

Question 1 asks you to write a balanced chemical equation.

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Limiting Reactants

Categorize

Options (3)

$$H_{2\left(g\right)}$$  

$$O_{2\left(g\right)}$$  

$$H_2O_{\left(g\right)}$$  

The reaction between hydrogen gas and oxygen gas creates water vapor. Organize the substances involved in the reaction.

Reactants

Products

Particle Diagram

Hydrogen and Oxygen are both diatomic.

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

​O

​H

​H

​O

Particle Diagram

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

​O

​H

​H

Multiple Choice

Is this the balanced chemical equation for hydrogen gas reacting with oxygen gas to produce water vapor:

$$2H_{2\left(g\right)}+O_{2\left(g\right)}\rightarrow\ 2H_2O_{\left(g\right)}$$

• Yes

• No

Multiple Choice

Is this an accurate particle diagram of the reaction?

• Yes

• No

Dropdown

​

Select answer

molecules of water can be produced. There is an excess of oxygen gas in the amount of ​

Select answer

molecules.

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

​Before:	​4	​	​4	​	​0
​Change:	​	​	​	​	​
​After:	​	​	​	​	​

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

​Before:	​4	​	​4	​	​0
​Change:	-4​	​	-2​	​	+4​
​After:	0​	​	2​	​	4​

You will need a piece of paper to take notes.

Watch the clip.

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​Let's step it up a notch.

Multiple Choice

What is the first step in using a BCA table?

• Make sure the chemical equation is balanced.

• Identify the given information.

• Determine what is needed to be found

• Find the mole ratios.

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.

Categorize

Options (2)

.118 mol

.180 mol

How many moles did you get?

Silver Nitrate

Calcium Chloride

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