Stoichiometry Module: ICE Tables
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Step 5: Complete the table.

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 final step is to complete the table. At this point, you should have the initial amount and the end amount for at least one reactant or product. From this you can calculate the change:

End amount (moles) - Initial amount (moles) = Change (moles)

It is important to realize that the change for the reactants should be negative since they are decreasing in amount while the change for the products should be positive since they are increasing in amount.

Once the change for one reactant or product is known, the change for the remaining reactants and products can be calculated by using the appropriate stoichiometric factor.

For example, the change for sodium carbonate is -0.05 moles (another way of saying this is that 0.05 moles of the sodium carbonate reacts). The balanced chemical equation tells us that every mole of sodium carbonate reacts with two moles of nitric acid. The amount of nitric acid that reacts can be calculated using this information:

What is the change for sodium nitrate?

+ 0.050 moles -0.050 moles +0.10 moles -0.10 moles

The balanced chemical equation tells us that for every mole of sodium carbonate that reacts, two moles of sodium nitrate are produced. Try again.

The balanced chemical equation tells us that for every mole of sodium carbonate that reacts, two moles of sodium nitrate are produced. Also remember that since the amount of sodium nitrate is increasing, the change would be positive. Try again.

Good! The balanced chemical equation tells us that for every mole of sodium carbonate that reacts, two moles of sodium nitrate are produced:

Once the change for all substances are known, the End amounts (moles) for all reactants and products can be calculated. Finally, the End amounts (moles) can be converted to units of mass or concentration (depending on what the problem asks for).

The completed ICE Table would look like:

Na2CO3 + 2 HNO3 2 NaNO3 + CO2 + H2O
Initial amount 5.3 g 250.0 mL 0 0 0
Initial (moles) .05 mol .125 mol 0 0 0
Change (moles) - 0.05 mol - 0.10 mol + 0.10 mol + 0.05 mol + 0.05 mol
End (moles) 0 mol .025 mol 0.10 mol 0.05 mol 0.05 mol
End amount 8.5 g

Hence, 8.5 grams of sodium nitrate are produced when 5.3 g sodium carbonate and 250.0 mL 0.5 M nitric acid are mixed.

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Remember that since the amount of sodium nitrate is increasing, the change would be positive. Try again.