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# Percent Yield

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Sections
Empirical Formula
Molecular Formula
Calculating Molar Mass
Mole Concept
Mass Percent
Stoichiometry
Limiting Reagent
Percent Yield

The Percent Yield determines how successful the product yield is in a chemical reaction.

###### Percent Yield

Concept #1: Percent Yield

Example #1: Consider the following balanced chemical reaction:
2 C6H6 (l) + 15 O2 (g) → 12 CO2 (g) + 6 H2O (l)

If a 2.6 g sample of C6H6 reacted with excess O2 to produce 1.25 g of water, what is the percent yield of water?

Practice: What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO4?

Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)

Practice: Ammonia, NH3, reacts with hypochlorite ion, OCl, to produce hydrazine, N2H4. How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has a 81.5% yield?

2 NH3 + OCl → N24 + Cl + H2O

Practice: The reduction of iron (III) oxide creates the following reaction:

Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g)

If the above reaction only went to 75% completion, how many moles of Fe2O3 were require to produce 0.850 moles of Fe?