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Reactions Involving a LIMITING REACTANT

Reactions Involving a LIMITING REACTANT. In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed. 2 NO(g) + O 2 (g) 2 NO 2 (g). LIMITING REACTANTS. R eactants.

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Reactions Involving a LIMITING REACTANT

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  1. Reactions Involving aLIMITING REACTANT • In a given reaction, there is not enough of one reagent to use up the other reagent completely. • The reagent in short supply LIMITS the quantity of product that can be formed.

  2. 2 NO(g) + O2 (g) 2 NO2(g) LIMITING REACTANTS Reactants Products Limiting reactant = ___________ Excess reactant = ____________

  3. LIMITING REACTANTS Demo of limiting reactants on Screen 4.7

  4. 1 2 3 LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 (See CD Screen 4.8) Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up 0.100 mol HCl

  5. LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl2 + H2 Rxn 1Rxn 2Rxn 3 mass Zn (g) 7.00 3.27 1.31 mol Zn 0.1070.0500.020 mol HCl 0.1000.1000.100 mol HCl/mol Zn 0.93/12.00/15.00/1 Lim Reactant LR = HCl no LR LR = Zn

  6. Reaction to be Studied 2 Al + 3 Cl2 ---> Al2Cl6

  7. Mass product Mass reactant Moles reactant Moles product PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form? Stoichiometric factor

  8. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

  9. Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio. 2 Al + 3 Cl2 ---> Al2Cl6 Reactants must be in the mole ratio

  10. Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 If There is not enough Al to use up all the Cl2 Lim reag = Al

  11. Deciding on the Limiting Reactant 2 Al + 3 Cl2 ---> Al2Cl6 If There is not enough Cl2 to use up all the Al Lim reag = Cl2

  12. Step 2 of LR problem: Calculate moles of each reactant We have 5.40 g of Al and 8.10 g of Cl2

  13. This should be 3/2 or 1.5/1 if reactants are present in the exact stoichiometric ratio. Find mole ratio of reactants 2 Al + 3 Cl2 ---> Al2Cl6 Limiting reagent is Cl2

  14. grams Cl2 grams Al2Cl6 moles Cl2 moles Al2Cl6 2 Al + 3 Cl2 ---> Al2Cl6 Limiting reactant = Cl2 Base all calcs. on Cl2 Mix 5.40 g of Al with 8.10 g of Cl2. What mass of Al2Cl6 can form?

  15. CALCULATIONS: calculate mass of Al2Cl6 expected. Step 1: Calculate moles of Al2Cl6 expected based on LR. Step 2: Calculate mass of Al2Cl6 expected based on LR.

  16. How much of which reactant will remain when reaction is complete? • Cl2 was the limiting reactant. Therefore, Al was present in excess. But how much? • First find how much Al was required. • Then find how much Al is in excess.

  17. Calculating Excess Al 2 Al + 3 Cl2 products 0.114 mol = LR 0.200 mol Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

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