# Limiting Reactant The coefficients in a balanced chemical reaction are a set ratio of products to products and products to reactants Much like a recipe.

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Limiting Reactant The coefficients in a balanced chemical reaction are a set ratio of products to products and products to reactants Much like a recipe calls for a set amount of each ingredient, a chemical reaction requires a set amount of each reactant. Recall that these ratios are in moles of reactant and product. If one of the reactants is insufficient in availability, it is called the limiting reactant because the reaction will stop when it runs out.

Finding Limiting Reactant Use for example the synthesis of ammonia from its elements 3H 2 (g) + 1N 2 (g) → 2NH 3 (g) 3 times (in moles) as much hydrogen is needed as compared to nitrogen. If you started out with equal amounts of each reactant, the hydrogen would run out first because more moles of it are required in the process. The hydrogen would then be the limiting reactant and the nitrogen would be the excess reactant

Find the Limiting Reactant Zn(s) + 2HCl(aq) → ZnCl 2 (aq) + H 2 (g) You have 5.0 moles of each zinc and hydrochloric acid. You have 7.5 moles of zinc and 3.0 moles of hydrochloric acid. You have 8.0 moles of hydrochloric acid and 3.5 moles of zinc. In each of these easy examples you can logically find the limiting reactant. For more difficult cases a set procedure is used.

Rules for finding the Limiting Reactant Calculate the amount of any product formed from the first reactant. Find the amount of the same product formed from the second reactant. The reactant that produces the smallest amount of product is the limiting reactant. The other reactant is in excess. The amount of product formed by the limiting reactant is called the theoretical yield. The amounts calculated by the other reactants are not valid because the limiting reactant would have run out.

Example #2 When aluminum is placed in a copper solution, the copper is oxidized and falls out of the solution 2Al (s) + 3CuSO 4 (aq) → 3Cu (s) + Al 2 (SO 4 ) 3 (aq) If 50.0 grams of each reactant is available, what is the theoretical yield and limiting reactant. Solve for the amount of Cu formed by each 50.0 g Al (1/26.98g)(3Cu/2Al)(63.55g/1) = 177g 50.0 g CuSO 4 (1/159.62)(3Cu/3CuSO 4 )(63.55g/1) = 19.9 g Since the copper(II) sulfate only produces 19.9 g of copper as compared to 177 g for the aluminum, the copper(II) sulfate is the limiting reactant and the 19.9 g of copper produced is the theoretical yield.

Calculating Actual Yield No reaction is 100% efficient. The actual yield is always going to be less than the calculated theoretical yield. The actual yield is the term used for the actual amount of product formed when the reaction is carried out. The percent yield is the ratio of the actual yield to theoretical yield times 100. % Yield = (actual/theoretical) * 100

Homework Pg 312-313 [27,28,34a,35a, 37,38]

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