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After leaning about moles, balanced chemical equations, unit conversions (dimensional analysis), Molar Mass Conversions, and simple stoichiometry, the concept of limiting reactants can be understood. Limiting reactant problems are at the heart of many chemical questions and a beginning student generally sees them specifically applied at least three times during General Chemistry in sections dealing with stoichiometry, molarity, and gases.
A limiting reactant is the reactant that limits the amount of product that can be formed.
In the simple stoichiometry section the examples had reactants in the same mole ratio as required by the balanced chemical equation. If the balanced equation had reactants in a 1:1 ratio, the grams were chosen to give the reactants a 1:1 mole ratio. What if the balanced equation had reactants in a 1:1 ratio, but the grams corresponded to a 2:1 ratio? In that case one of the reactants would be completely used up and there would be extra of the other reactant. The reactant that is all used up is the limiting reactant. The amount of limiting reactant determines the amount of product that can be formed.
The grams of reactants can be converted to moles of reactants using the conversion factors found from the periodic table. The limiting reactant can be determined using the mole ratio in the balanced chemical equation. The moles of the limiting reactant can then be converted to the moles of product using the mole ratios in the balanced chemical equation. Finally, the moles of product can then be converted to grams of product using the conversion factors found from the periodic table for the products. This procedure is just like the simple stoichiometry procedure, except the limiting reactant must be determined before calculating how much product is formed.
In this section you will learn to determine which reactant is the limiting reactant and then use the limiting reactant to determine the amount of product formed.
Summary of Procedure Starting With Grams
- Convert initial grams to initial moles. Use the Periodic Table to get conversion factors.
- Determine the limiting reactant. Use the balanced chemical equation to get conversion factors.
- Use the limiting reactant to determine the moles of product formed. Get conversion factors from the balanced chemical equation.
- Convert moles of product to grams of product. Use the Periodic Table to get conversion factors.
Example #1 - Starting With Moles
Consider the following representation of the chemical reaction between hydrogen and oxygen:
2H2(g) + O2(g) → 2H2O(g)
The balanced chemical equation can be interpreted as two moles of hydrogen gas react with one mole of oxygen gas to form two moles of water.
If it is known that two moles of hydrogen gas and two mole of oxygen gas are initially present, the actual mole ratios will not conform to those required by the balanced chemical equation and one of the reactants will be a limiting reactant. The limiting reactant can be determined by choosing one of the reactants and determining how much of the other reactant is needed in order for complete reaction of the first one to take place. In this case you could choose hydrogen and determine how much oxygen is needed to completely react with all of the hydrogen. Or you could choose oxygen and determine how much hydrogen is needed to completely react all of the oxygen. In either case you can see if the amount needed is more or less than the amount that you actually have and from that decide which reactant is the limiting reactant.
For the example of two moles of hydrogen gas reacting with two moles of oxygen gas, you could pick hydrogen and determine the amount of oxygen needed. Use the mole ratios from the balanced chemical equation to make the conversion from moles of hydrogen to moles of oxygen.
O2 needed for complete reaction of the hydrogen:
( 2 mole H2 ) (
1 mole O2O 2 mole H2 ) = 1 mole O2 needed There are actually two moles of oxygen available, which means that there is extra oxygen. This also means that the hydrogen will all be used up and hydrogen is the limiting reactant.
What if you would have chosen the oxygen? Again use the mole ratios from the balanced chemical equation.
H2 needed for complete reaction of the oxygen:
( 2 mole O2 ) (
2 mole H2 1 mole O2 ) = 4 mole H2 needed There are actually two moles of H2, which isn't enough to react with all of the oxygen, and so hydrogen is again determined to be the limiting reactant.
Once the limiting reactant is determined, the moles of product can be determined. Use the mole ratios from the balanced chemical equation to convert from moles of limiting reactant to moles of product.
( 2 mole H2 ) (
2 mole H2O 2 mole H2 ) = 2 mole H2O Two moles of H2O can be produced from two moles of hydrogen and two moles of oxygen.
Example #2 - Starting With Grams
What if you started with 10 g of H2 and 10 g of O2, how many grams of water would be produced in that case?
Convert initial grams of reactants to initial moles of reactants.
Moles of H2 from grams of H2:
( 10 g H2 ) (
1 mole H2 2 g H2 ) = 5 mole H2 Moles of O2 from grams of O2:
Determine limiting reactant.
( 10 g O2 ) (
1 mole O2 32 g O2 ) = 0.31 mole O2 O2 needed to react all of the H2:
( 5 mole H2 ) (
1 mole O2 2 mole H2 ) = 2.5 mole O2 needed There are actually only 0.31 moles of O2 available, which means that O2 is the limiting reactant.
Use the limiting reactant to determine the amount of product produced (moles and then grams). Use the mole ratio from the balanced equation to convert to moles of water and the molar mass, calculated using the periodic table, to convert to grams of water.
( 0.31 mole O2 ) (
2 mole H2O 1 mole O2 ) (
18 g H2O 1 mole H2O ) = 11.6 g H2O The maximum grams of water that can be produced from the 10 g samples is 11.6 g H2O.
Example #3 - Complete Example
Question: How many grams of NaCl will be produced when 30.0 g Na reacts with 40.0 g Cl2?
Write the balanced equation:
2Na + Cl2 → 2NaCl
Convert grams of reactants to moles of reactants:
Moles of Na from grams of Na:
( 30 g Na ) (
1 mole Na 23 g Na ) = 1.3 mole Na Moles of Cl2 from grams of Cl2:
( 40 g Cl2 ) (
1 mole Cl2 71 g Cl2 ) = 0.56 mole Cl2 Determine the limiting reactant.
Cl2 needed to react all of the Na:
( 1.3 mole Na ) (
1 mole Cl2 2 mole Na ) = 0.65 mole Cl2 needed Since 0.65 mole Cl2 are needed and there are only 0.56 mole Cl2 available, Cl2 will be the limiting reactant.
Determine moles and then grams of product (NaCl):
( 0.56 mole Cl2 ) (
2 mole NaCl 1 mole Cl2 ) (
58.5 g NaCl 1 mole NaCl ) = 65.5 g NaCl A maximum of 65.5 g NaCl can be produced in this case.