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Stoichiometry of Excess-Limiting Reactions

Stoichiometry of Excess-Limiting Reactions. A = B = C =. +. ?. +. 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”. Excess-Limiting Concept.

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Stoichiometry of Excess-Limiting Reactions

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  1. StoichiometryofExcess-Limiting Reactions

  2. A = B = C = + ? + 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”. 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?

  3. 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. 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. 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. In other words, it takes 3 times as many Mg atoms as it does N2 molecules.

  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 ? 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? It’s important because if we have equal masses of these two, we will actually have more Mg atoms than N2 molecules since each Mg is lighter in mass.

  6. 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 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. 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”.

  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 ? To react all of the 10.g of Mg that we “have”, we “need” 3.84g of N2. We already “have” 5.0 g of N2 so since we only “need” 3.84g of N2,we have more than we need. 5.0g of N2 – 3.84g N2 needed leaves 1.16 g excess N2.

  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 The 1.16 g of excess N2 will not react because there is not enough Mg to react with it. Since there will be left over N2 that is un-reacted, we call the N2 the excess reactant. 5.0g of N2 – 3.84g N2 needed leaves 1.16 g excess N2.

  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 ? Excess 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 N2. When the 10. g of Mg is used up, the reaction will stop.

  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 With all of the Mg gone, no more N2 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.

  12. 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

  13. 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.

  14. Answer Excess Al2O3 + 6 HF  2 AlF3 + 3 H2O HAVE NEED Limiting

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