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Limiting Reagents

Limiting Reagents. Explaining 3 methods for finding the limiting reagent in a chemical reaction. Limiting Reagents. Limiting reagents are very important in chemical calculations as it is the limiting reagent that determines the quantity of products that you get from chemical reactions.

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Limiting Reagents

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  1. Limiting Reagents Explaining 3 methods for finding the limiting reagent in a chemical reaction. 2AB Chemistry

  2. Limiting Reagents • Limiting reagents are very important in chemical calculations as it is the limiting reagent that determines the quantity of products that you get from chemical reactions. • In previous stoichiometry problems you have been given enough information to find the number of moles of one reagent. • The other reagent was probably describes as being ‘sufficient’. This means that there was plenty of it, or there was excess of this reagent. • In limiting reagent problems you will be provided with enough information to calculate the moles of both reactants 2AB Chemistry

  3. Method 1: Logical Reasoning • This method is described as Method 1 in the 2AB Text book Chemistry for WA 1 (2011) p284. • Choose one reactant and calculate how much of the other reactant you would need to complete the reaction. • Then compare how much of that reactant you have to how much you need. • The logical reasoning is that if you have less than you need then this reagent is limiting. • The other reagent must therefore be in excess. 2AB Chemistry

  4. Method 1: Example • A mixture of 4 moles of hydrogen gas and 3 moles of oxygen gas react to form water. Determine the limiting reagent. • Step 1: write the balanced chemical reaction. • Step 2: decide on one reagent to be your knownand calculate how much of the other reagent you will need. 2AB Chemistry

  5. Method 1: Example continued • A mixture of 4 moles of hydrogen gas and 3 moles of oxygen gas react to form water. Determine the limiting reagent. • Step 3: compare the moles of reagent you need to the moles of reagent you have. • I need 2 moles of O2, but I have 3 moles of O2. • Step 4: Make the logical statement. • Therefore O2 is in excess and H2must be the limiting reagent. 2AB Chemistry

  6. Method 2: Comparing Ratios • This method is described as Method 2 in the 2AB Text book Chemistry for WA 1 (2011) p284. With a modification to make it easier. • From the balanced reaction write the stoichiometric ratio of the reactants. • From the information in the question write the actual ratio of the reactants. • Compare the two ratios. • The logical reasoning is that if the actual ratio is less that the stoichiometric ratio then the reagent on top (numerator) is too small or you have less than you need then this reagent is limiting. • The other reagent must therefore be in excess. 2AB Chemistry

  7. Method 2: Comparing Ratios continued • It follows that if the actual ratio is greater that the stoichiometric ratio then the reagent on top (numerator) is too big or you have morethan you need then this reagent is in excess. • The other reagent must therefore be limiting. 2AB Chemistry

  8. Method 2: Example • A mixture of 3 moles of hydrochloric acid and 5 moles of zinc metal react to form a salt and water. Determine the limiting reagent. • Step 1: write the balanced chemical reaction. • Step 2: write the stoichiometric ratio and actual ratio. 2AB Chemistry

  9. Method 2: Example continued • A mixture of 3 moles of hydrochloric acid and 5 moles of zinc metal react to form a salt and water. Determine the limiting reagent. • Step 3 compare the ratios. • The actual ratio is less than the stoichiometric ratio. • Step 4: Make the logical statement. • Therefore HCl is limiting as the numerator is too small and therefore it follows that Zn must be in excess. 2AB Chemistry

  10. Method 2: Comparing Ratios Modification • A different way to compare the ratios is to convert the actual ratio to a ratio in which the denominator is the same as the denominator of the stoichiometric ratio. • The value of the actual numerator can be compared to the value of the stoichiometric numerator. 2AB Chemistry

  11. Method 2: Example Modified • A mixture of 3 moles of hydrochloric acid and 5 moles of zinc metal react to form a salt and water. Determine the limiting reagent. • Step 2a: write the stoichiometric ratio and actual ratio. And convert the actual ratio. 2AB Chemistry

  12. Method 2: Example Modified continued • A mixture of 3 moles of hydrochloric acid and 5 moles of zinc metal react to form a salt and water. Determine the limiting reagent. • Step 3a: compare the numerators. • The actual numerator is less than the stoichiometric numerator. • Step 4a: Make the logical statement. • Therefore as HCl is the numerator it is limiting as the actual numerator is smaller than the stoichiometric numerator and therefore it follows that Zn must be in excess. 2AB Chemistry

  13. Method 3: Moles in Holes • This method is a modified version of Method 1, where the number of moles needed for each reactant is found and compared to the actual number of moles. • Write the balanced chemical reaction. • Place two rows of boxes (holes) below each reactant with the first row labeled ‘have’ and the second row labeled ‘need’. • Place the number of moles of each reactant in the ‘have’boxes below each reactant. • Choose one reactant and calculate how much of the other reactant you would need to complete the reaction. Place this answer in the ‘need’ box. • Then do the same for the other reactant. 2AB Chemistry

  14. Method 3: Moles in Holes continued • Now compare how much you have to how much you need. • One column will show that you have less of that reactant than you need. This will be the limiting reagent. • The other column will show that you have more than you need. This will be the excess reagent. 2AB Chemistry

  15. Method 3: Example • A mixture of 5 moles of aluminium metal and 3 moles of chlorine gas react to form aluminium chloride. Determine the limiting reagent. • Step 1: write the balanced chemical reaction. • Step 2: Set up the holes for your moles and fill in the ‘have’ holes 2AB Chemistry

  16. Method 3: Example continued • A mixture of 5 moles of aluminium metal and 3 moles of chlorine gas react to form aluminium chloride. Determine the limiting reagent. • Step 3: Calculate the number of moles of each reactant that you need. 2AB Chemistry

  17. Method 3: Example continued • A mixture of 5 moles of aluminium metal and 3 moles of chlorine gas react to form aluminium chloride. Determine the limiting reagent. • Step 4: Place these values in the ‘need’ holes. 2AB Chemistry

  18. Method 3: Example continued • A mixture of 5 moles of aluminium metal and 3 moles of chlorine gas react to form aluminium chloride. Determine the limiting reagent. • Step 5: compare the moles of reagent you need to the moles of reagent you have. • I need 2 moles of Al, but I have 5 moles of Al. • I need 7.5 moles of Cl2, but I only have 3. • Step 6: Make the logical statement. • Therefore Al is in excess and Cl2 must be the limiting reagent. 2AB Chemistry

  19. Method 3: Moles in Holes continued • Be careful with his method, as once moles start digging holes it is hard to stop them! • Holes can be dug for ‘used’ and ‘remain’. LR 2AB Chemistry

  20. Method 4: Mr Fry’s Method • This method is a modified version of the Modified Method 2, where the actual mole ratio is converted to a 1 : 1 ratio. • Write the balanced chemical reaction. • Write the actual ratio of moles. • Divide the actual number of moles by the stoichiometric number of moles. • The smallest number is the limiting reagent. 2AB Chemistry

  21. Method 4: Example • A mixture of 0.145 moles of aluminum and 0.375 moles of chlorine gasreact to form a salt. Determine the limiting reagent. • Step 1: write the balanced chemical equation • Step 2: write the actual number of moles and divide by the stoichiometric number of moles. 2AB Chemistry

  22. Method 3: Example continued • A mixture of 0.145 moles of aluminum and 0.375 moles of chlorine gas react to form a salt. Determine the limiting reagent. • Step 3: compare the moles of reagent, the smallest is the limiting reagent. • 0.0725 moles of Al is less than 0.125 moles of Cl2. • Step 4: Make the logical statement. • For every 0.125 moles of Cl2 I need 0.125 moles of Al, but only have 0.0725 moles of Al therefore Al is the limiting reagent. 2AB Chemistry

  23. Practice: Limiting Reagent • Use Method 1 to answer this question. • 1. Determine the limiting reagent if 2.5 moles of Al2(CO3) 3 reacts with 3.0 moles of HCl. • Use either of Method 2 to answer this question. • 2. Determine the limiting reagent if 3.0 moles of C2H6reacts with 7.5 moles of O2. • Use Method 3 to answer this question. • 3. Determine the limiting reagent if 10.5 moles of S reacts with 5.0 moles of O2. • Use Method 4 to answer this question. • 4. Determine the limiting reagent if 0.55 moles of C6H12reacts with 0.30 moles of O2. 2AB Chemistry

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