EXCESS AND LIMITING REAGENTS!! PRACTICE MAKES PERFECT.

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Presentation transcript:

EXCESS AND LIMITING REAGENTS!! PRACTICE MAKES PERFECT

WHAT YOU WILL NEED : 1) Knowledge on stoichiometry (featured on this website)

2) Periodic table

3) Understanding of significant figures (featured on this website)

4) To know how to balance equations (featured on this website)

WHAT IS EXCESS AND LIMITING REAGENTS? The limiting reagent is the substance that is totally consumed or used up when the chemical reaction is complete. It is called limiting because there is not enough of it, it is limiting. There is never extra left. On the other hand, the excess reagent is the substance that contains a greater amount than necessary to react completely with the limiting reagent. When the chemical equation is complete, there is an excess of that substance since it had more than enough of what was required to complete the chemical equation.

LIMITING REAGENTS EXAMPLE I have 500 nails and 100 pieces of wood, this makes one house. There are 2000 nails and 200 pieces of wood, I want to make 3 houses, what is the limiting and excess reagent? The limiting reagent would be the wood because there is missing 100 pieces and the excess reagent would be the nails because I have an extra of 500 nails.

THE METHOD. To solve a question of which reactant is the limiting reagent (given the initial amount for each reactant) is to calculate the amount of product that could be made from each amount of reactant, assuming all other reactants are available in unlimited amounts. In this case, the limiting reagent will be the one that produces the least amount of product.

EXAMPLE 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) We have 2.00 grams of NH 3 and 4.00 grams O. What is the limiting reagent and how much excess is remaining? NEXT: Use stoichimoetry to determine how much product is formed by each reactant. 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (g) We have 2.00 grams of NH 3 and 4.00 grams O. What is the limiting reagent and how much excess is remaining? NEXT: Use stoichimoetry to determine how much product is formed by each reactant. ANS: In this case, the oxygen produces the lesser amount of product, therefore it is the limiting reagent!!

Next, to find the amount of excess reactant, we need to calculate how much of the non-limiting reactant (ammonia) actually did react with the limiting reactant (oxygen), because there is going to be extra left since it is the excess. We're not finished yet!!! 1.70 g is the amount of ammonia that reacted, not what is left over. To find the amount of excess reactant left over, subtract the amount that reacted from the amount in the original sample. 2.00g NH 3 (original) – 1.70g NH 3 (reacted) = 0.30g NH3 remaining ANS: 0.30 g NH 3 remaining

BIBLIOGRAPHY Capraro, Ernest. "Helium." Finding the Limiting Reactant (n.d.): 1-2. Web. 11 Oct Teachers class notes, Chemistry class. (Ms. Young)