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Calculating Limiting and Excess Reagents Section 5.5 SCH 3U.

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Presentation on theme: "Calculating Limiting and Excess Reagents Section 5.5 SCH 3U."— Presentation transcript:

1 Calculating Limiting and Excess Reagents Section 5.5 SCH 3U

2 The Mole Ratio The quantities of substances that react in a chemical reaction are determined by the mole ratio given by the balanced chemical equation but, in the real world, it is less likely for the reactants to be in the mole ratios given by the equation – WHY??

3 The Answer In nature, compounds are found in random quantities In labs / industry, we might want to ensure that one reactant is completely used up

4 In any chemical reaction, one reactant is always present in EXCESS - in other words: the reagent that is not completely used up in the reaction there will be some left over at the end of the reaction in order to provide a reagent in excess, chemists add 10% to the mass required The Excess Reagent

5 The other reactant is LIMITING - in other words: it is completely used up in the reaction the reagent in a chemical reaction that limits the amount of product that is able to form The Limiting Reagent

6 Kitchen Stoichiometry – an analogy A recipe to make 30 candies requires the following: 2 cups water 5 cups sugar 1 cup strawberries Equation 2 c water + 5 c sugar + 1 c strawberries  30 candies

7 a. If you were given 10 cups of water, 10 cups of sugar and 10 cups of strawberries, which ingredient would limit the amount of candies being made? Which reagent is in excess? Questions:

8 b. How many candies can you make with the ingredients given in part a? Questions:

9 c. How much of the excess reagents will be left over? Questions:

10 Lithium nitride reacts with water to produce ammonia gas and lithium hydroxide. If 4.5 g of lithium nitride and 5.8 g of water is available: a. How much ammonia gas (in grams) can be produced? Step 1: Write a balanced chemical equation. Step 2: Choose a product – pick the one that you will need to calculate the amount of (ammonia). Step 3: Determine the amount of ammonia that can be produced from each of the reactants. Step 4: Compare the amounts of product formed from the lithium nitride and from the water – the one that produces the least ammonia is the limiting reagent Example 1:

11 Lithium nitride reacts with water to produce ammonia gas and lithium hydroxide. If 4.5 g of lithium nitride and 5.8 g of water is available: b. How much (in grams) of the excess reagent will be left over? Li 3 N (s) + 3H 2 O (l) → 3LiOH (aq) + NH 3(g) Example 1:

12 Nitrogen and hydrogen react to form ammonia according to the following equation: N 2(aq) + H 2(g) → NH 3(g) If g of N 2 is to be completely used up, what is a reasonable mass of H 2 to use in the reaction? Example 2:


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