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Go to the Molecules mode by selecting the icon at the bottom. Notice that there are three real reactions: the synthesis of water, the synthesis of ammonia, and the combustion of methane. 1)Select each of the reactions and observe the reactants taking part in the reactions. 2)Suppose you are carrying out each of these reactions starting with five moles of each reactant, observe that some reactants are in excess whereas some reactants limit the amount of products formed. 3)Classify each of the reactants as a limiting reactant or an excess reactant for a reaction starting with five moles of each reactant. 4)Drag the appropriate items to their respective bins. View Available Hint(s) ResetHelp.

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Answer:

Synthesis of water: [tex]H_{2}[/tex] is limiting reactant ; [tex]O_{2}[/tex] is excess reactant

Synthesis of ammonia: [tex]H_{2}[/tex] is limiting reactantt ; [tex]N_{2}[/tex] is excess reactant

Combustion of methane: [tex]O_{2}[/tex] is limiting reactant ; [tex]CH_{4}[/tex] is excess reactant

Explanation:

In a chemical reaction, if a reactant is totally consumed when the reaction is completed,it is called the limiting reactant/reagent , while the other reactant is called the excess reactant/reagent. The product formed is limited because the reaction stops abruptly as one of the reactant is consumed completely.

In order to determine which is the limiting reactant in a particular chemical reaction:

  1. Calculate the amount of product formed if the first reactant were completely consumed
  2. Repeat this calculation for the second reactant.
  3. The reactant that produces the smaller amount of product is the limiting reactant.
  • Synthesis of water

         [tex]2H_{2}[/tex] + [tex]O_{2}[/tex] → [tex]2H_{2}O[/tex]

[tex]H_{2} : H_{2}O[/tex]

2  → 2

5  → x

x= 5*2/2

x= 5 moles of water

[tex]O_{2} : H_{2}O[/tex]

1   → 2

5  → x

x= 5*2/1

x= 10 moles

limiting reactant: [tex]H_{2}[/tex]

excess reactant: [tex]O_{2}[/tex]

  • Synthesis of ammonia

       [tex]N_{2} + 3H_{2}[/tex] → [tex]2NH_{3}[/tex]

[tex]N_{2}[/tex] → [tex]NH_{3}[/tex]

1    →  2

5   → x

x= 5*2/1

x= 10 moles ammonia

[tex]H_{2}[/tex] → [tex]NH_{3}[/tex]

3   → 2

5  → x

x= 5*2/3

x= 3.33 moles of ammonia

limiting reactant: [tex]H_{2}[/tex]

excess reactant: [tex]N_{2}[/tex]

  • Combustion of methane

        [tex]CH_{4} + 2O_{2}[/tex] → [tex]CO_{2} + 2H_{2}O[/tex]

[tex]CH_{4}[/tex] → [tex]CO_{2}[/tex]

1      →   1

5    →    x

x= 5*1 /1

x= 5 moles of [tex]CO_{2}[/tex]

[tex]O_{2}[/tex] → [tex]CO_{2}[/tex]

2  →  1

5  →  x

x= 5*1 /2

x= 2.5 moles of [tex]CO_{2}[/tex]

limiting reagent: [tex]O_{2}[/tex]

excess reagent:  [tex]CH_{4}[/tex]

     

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