Step 4

	Stoichiometry Module: ICE Tables

Table of Contents

General Stoichiometry

Limiting Reactants

Chemical Analysis

Step 4: Determine the Change (moles) amount for one substance in the reaction.

How many grams of NaNO₃ are produced when 5.3 grams of Na₂CO₃ are added to 250.0 mL of 0.50 M HNO₃ and the reaction is allowed to go to completion?

Na₂CO₃ (aq)

+

2 HNO₃(aq)

2 NaNO₃ (aq)

+

CO₂ (g)

+

H₂O (l)

The next step is to decide the change in moles for one reactant or product. There are two possibilities:

The reaction goes to completion. In this case, the reaction will continue until one of the reactants is gone. You must determine which reactant is limiting.
The reaction does not go to completion. In this case, the amount of one of the reactants that actually is used in the reaction or the amount of products that are formed would be given.

In this example, the reaction goes to completion.

Which reactant limits the reaction?

Na₂CO₃	HNO₃

Good! The 0.05 moles Na₂CO₃ initially present would require 0.10 moles HNO₃to react completely:

There is more HNO₃than 0.10 moles so all 0.050 moles of the Na₂CO₃ can react.

Since the sodium carbonate limits the reaction, it will react completely (or none will be present at the end of the reaction.

The ICE Table with this information looks like:

Change (moles)
	Na₂CO₃	+	2 HNO₃		2 NaNO₃	+	CO₂	+	H₂O
Initial amount	5.3 g		250.0 mL		0		0		0
Initial (moles)	.05 moles		.125 moles		0		0		0
End (moles)	0 moles
End amount

The 0.125 moles HNO₃initially present would require 0.0625 moles Na₂CO₃ to react completely:

There is only 0.050 moles Na₂CO₃O₃initially - not enough to react with all of the nitric acid.

Since the sodium carbonate limits the reaction, it will react completely (or none will be present at the end of the reaction.

The ICE Table with this information looks like:

Change (moles)
	Na₂CO₃	+	2 HNO₃		2 NaNO₃	+	CO₂	+	H₂O
Initial amount	5.3 g		250.0 mL		0		0		0
Initial (moles)	.05 moles		.125 moles		0		0		0
End (moles)	0 moles
End amount