CR - Stoichiometry (Lesson)

Stoichiometry

Stoichiometry is the study of the relationships between relative quantities of substances taking part in a chemical reaction. Writing a balanced chemical equation is required by the Law of Conservation of Mass, and those coefficients are used as mole ratios to calculate quantities of substances in the reaction. 

Stoichiometry can be used to answer various types of questions...

• How much product will this reaction make?
• How much of a reactant will I need in order to make a certain amount of product?
• Which reactant will I run out of first?
• If two reactions make the same product, which one would be more efficient and require fewer products to get the same amount of product?

Need a quick review? If so, watch the video below. Be sure your volume is turned on!

You Try It: Stoichiometry

Limiting Reactants

Even though there are two reactants, the stoichiometry examples done so far have only mentioned one reactant. Often a problem will state that the other reactant is in excess. Sometimes, that is just assumed. But, what do we do if a stoichiometry problem mentions more than one reactant amount?

Use the following equation to help you work through the following limiting reactants practice problem:

2 C₆H₁₀ + 17 O₂ $\longrightarrow$ 12 CO₂ + 10 H₂O

Practice Problem One: If there are 35.0 grams C₆H₁₀ and 45.0 grams of O₂, what is the limiting reactant (i.e. which reactant will be used up first)?

Step One: Convert each substance to moles.

C₆H₁₀: 35.0 g / 82.145 g/mol = 0.426 mol

O₂: 45.0 g / 31.998 g/mol - 1.406 mol

Step Two: Divide by coefficients

C₆H₁₀: 0.426 mol/2= 0.213

O₂: 1.406 mol/17 = 0.083

O₂ is the limiting reactant. 

Practice Problem Two: Determine how many grams of the excess reagent are used up when the limiting reagent is fully consumed.

Practice Problem Three: Determine grams of C₆H₁₀  are remaining by subtraction of starting - used for the reactants.

35.0 – 13.6 g = 21.4 g  C₆H₁₀ remaining

Percent Yield

The amount of product that may be produced by a reaction under specified conditions, as calculated per the stoichiometry of an appropriate balanced chemical equation, is called the theoretical yield of the reaction.

An experimental amount that is actually obtained is known as the actual yield and is often less than the theoretical yield for a number of reasons.

• Some reactions are inherently inefficient, being accompanied by side reactions that generate other products.
• Others are, by nature, incomplete and never react to 100% completion.
• Some products are difficult to collect without some loss, and so less than perfect recovery will reduce the actual yield.

The extent to which a reaction’s theoretical yield is achieved is commonly expressed as its percent yield:

Percent Yield = (actual yield)/(theoretical yield) x 100

Consider this 'non-chemistry' example. Let's say you take a quiz that is worth 25 points. The theoretical yield for this quiz would be 25 and represents the number of points you should get "in theory," the number of points possible. In reality, let's say that you missed a few questions and scored 21 points. This is your actual yield. The percent yield represents your grade. All you have to do is divide your actual grade by the grade you should have made "in theory", and then multiply by 100. That is how you know you make 84%!

Watch the video below to learn more about working stoichiometry and percent yield problems. Be sure your volume is turned on!

Let's work through the following problem:

A 50.6 g sample of Mg(OH)₂ reacts with 45.0 g of HCl.  A chemist in the lab actually obtained 55.4 g of MgCl₂. What is the percent yield of this reaction?

Step 1: Find the Limiting Reactant

Balanced equation: Mg(OH)₂ + 2 HCl $\longrightarrow$ MgCl₂ + 2 H₂O

Compare the two reactants in the mole ratio.

50.6g Mg(OH)₂ X 1 mol Mg(OH)₂/ 58.3 g x 2 mol HCl/1 mol Mg(OH)₂ x 36.5 g HCl /1 mol HCl = 63.36 grams HCl

Since 63.35 grams HCl is needed, but only 45.0 grams are available, HCl limits. 

Step 2: Find Theoretical Yield

Using the limiting reactant HCl, calculate the mass of MgCl₂ that can be produced. Be sure to use the amount available, not the needed amount to use up all the excess reactant.

45.0 grams HCl  x 1 mole HCl/36.5 g HCl x 1 mol MgCl₂/ 2 mol HCl x 95.3 g MgCl₂/ 1 mol MgCl₂ = 58.7 g MgCl₂

Step 3: Calculate Percent Yield. 

 (55.4 g MgCl₂  / 58.7 g MgCl₂ ) x 100 = 94.4 %

You Try It: Percent Yield

In the synthesis of chromium (III) chloride from the constituent elements, only 10.8 grams was produced from 4.1 grams of chromium metal and 9.2 grams of chlorine gas. Determine the percent yield of this reaction. Click on the plus signs to check your answers!

2Cr + 3Cl₂ $\longrightarrow$ 2CrCl₃

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