Excess-Limiting Concept

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

2. EXCESS A = B = C = ? + A + B  C Excess-Limiting Concept 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.

3. Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? Suppose we have 10. grams of Mg and 5.0 g of N2 for this reaction. How much product will we be able to make? We can quickly see that this reaction requires 3 moles of Mg to every 1 mole of N2. .

4. Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? The other issue is molar mass. Mg has a molar mass of 24.3 g/mol and the N2 has a molar mass of 28.0 g/mol. How does this relate?

5. Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? This is done by using the usual stoichiometric process of: grams A  moles A; moles A  moles B; moles B  grams B

6. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 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”.

7. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? 5.0g of N2 – 3.84g N2 needed leaves 1.16 g excess N2.

8. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? Excess

9. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? Excess

10. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? Limiting Excess With all of the Mg gone, no more N2 can be used and no more product can be made.

11. We “have” this much We “need” this much Let’s see how this works with an actual reaction. Excess-Limiting Consider the reaction:3 Mg + N2 Mg3N2 10. g 5.0 g ? Limiting Excess Now we can answer the original question which was how much product will we make? Using the Limiting

12. Have we learned it yet? Try this one on your own: Al2O3 + 6 HF  2 AlF3 + 3 H2O How many grams of H2O can be made with 150g of aluminum oxide and 150g of hydrofluoric acid? • Follow these steps: • Use the 150g of Al2O3 to find the mass of HF needed. • Use the 150 g of HF to find the mass of Al2O3 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 H2O • that will be produced.