1. Stoichiometry of Excess-Limiting Reactions

2. Excess-Limiting Concept Consider the simple reaction: A + B  C It means that 1 mole of “A” reacts with 1 mole of “B” and produces 1 mole of “C”. + A = B = C = ? ? + But, what if we actually put 2 moles of “A” into a container with only 1 mole of “B” ? There is one excess “A” left over when the reaction is finished. Why does this happen ? EXCESS

3. Excess-Limiting Concept A + B  C According to the equation, they MUST combine in a 1:1 ratio. After the 1 mole of B is all used up, we will have only used 1 mole of A and there will still be 1 mole of A left over. ? ? + EXCESS A = B = C = Because there is more than enough A for the reaction, A is called the excess reactant. When B is used up, the reaction stops. B is called the limiting reactant because it limits how much C the reaction can make.

4. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 Suppose we have 10. grams of Mg and 5.0 g of N 2 for this reaction. How much product will we be able to make? 10. g 5.0 g ? We can quickly see that this reaction requires 3 moles of Mg to every 1 mole of N 2. In other words, it takes 3 times as many Mg atoms as it does N 2 molecules.

5. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? The other issue is molar mass. Mg has a molar mass of 24.3 g/mol and the N 2 has a molar mass of 28.0 g/mol. How does this relate? It’s important because if we have equal masses of these two, we will actually have more Mg atoms than N 2 molecules since each Mg is lighter in mass.

6. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We need to identify the limiting reactant… the one that will be used up first, before we can determine the amount of product that can be made in this situation. This is done by using the usual stoichiometric process of: grams A  moles A; moles A  moles B; moles B  grams B

7. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We begin by using what we “have” of each reactant and calculate the amount of each that we would “need”. We “have” this much We “need” this much

8. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We “have” this much We “need” this much To react all of the 10.g of Mg that we “have”, we “need” 3.84g of N 2. 5.0g of N 2 – 3.84g N 2 needed leaves 1.16 g excess N 2. We already “have” 5.0 g of N 2 so since we only “need” 3.84g of N 2, we have more than we need.

9. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We “have” this much We “need” this much 5.0g of N 2 – 3.84g N 2 needed leaves 1.16 g excess N 2. The 1.16 g of excess N 2 will not react because there is not enough Mg to react with it. Since there will be left over N 2 that is un-reacted, we call the N 2 the excess reactant. Excess

10. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We “have” this much We “need” this much Comparing the 10. g of Mg that we “have” to the 13 g of Mg that we “need”, we see that we don’t “have” enough Mg to react with all of the N 2. When the 10. g of Mg is used up, the reaction will stop. Excess

11. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We “have” this much We “need” this much With all of the Mg gone, no more N 2 can be used and no more product can be made. Mg is called the limiting reactant because when it is gone, the reaction stops. It therefore “limits” the amount of product that can be made. Excess Limiting

12. Excess-Limiting Let’s see how this works with an actual reaction. Consider the reaction: 3 Mg + N 2  Mg 3 N 2 10. g 5.0 g ? We “have” this much We “need” this much Now we can answer the original question which was how much product will we make? Excess Limiting Using the Limiting

13. Have we learned it yet? Try this one on your own: Al 2 O 3 + 6 HF  2 AlF 3 + 3 H 2 O How many grams of H 2 O can be made with 150g of aluminum oxide and 150g of hydrofluoric acid? Follow these steps: Use the 150g of Al 2 O 3 to find the mass of HF needed. Use the 150 g of HF to find the mass of Al 2 O 3 needed. Compare the amounts you “have” to the amounts you “need” and determine the limiting reactant (have<need). Use the limiting reactant amount to calculate the mass of H 2 O that will be produced.

14. Al 2 O 3 + 6 HF  2 AlF 3 + 3 H 2 O Answer HAVE NEED Excess Limiting