Limiting Reactant Lab Answer Key

Learning Objectives

Apply stoichiometry to make up one's mind the limiting reactant.
Calculate the theoretical yield.
Calculate the pct yield of a reaction.

Introduction

In lecture yous have learned to read chemical equations and evaluate the mol to mol ratios of reactants and products involved in a chemical reaction. In laboratory experiments it is difficult to measure out chemicals in the exact ratio necessary for the chemical reaction. For time and speed reasons, the reaction mixtures in lab will unremarkably have a limiting and an backlog reactant.

Limiting reagent (also chosen limiting reactant) problems use stoichiometry to determine the theoretical yield for a chemical reaction. The limiting reactant volition be completely consumed in the reaction and limits the amount of product you lot tin make. The limiting reactant also determines the corporeality of production you lot can make (the theoretical yield). The reactant that is left over after the reaction is complete is chosen the excess reactant.

Instance ane

Consider the process information technology takes to make a ham sandwich. Yous need ii slices of bread and ane piece of ham to make each sandwich. How many complete sandwiches could you lot make if you had 18 slices of bread and six slices of ham? Allow's set up this upward like a chemical equation where the ham and staff of life are our reactants and the sandwich is our product:

2 slices of breadstuff + 1 piece of ham = i ham sandwich

Clip art of bread, ham, and a sandwich

Now nosotros can use stoichiometry to determine the amount of sandwiches nosotros could brand if nosotros used all of our reactants.

[latex]\displaystyle18\text{ slices of staff of life}\frac{ane\text{ sandwich}}{2 \text{ pieces of bread}}=9\text{ sandwiches}\\[/latex]

[latex]\displaystyle6\text{ slices of staff of life}\frac{1\text{ sandwich}}{2 \text{ piece of ham}}=6\text{ sandwiches}\\[/latex]

Nosotros accept enough bread to make 9 sandwiches. In that location is not enough ham bachelor to make as many sandwiches. If we use all the ham, we can only make half dozen sandwiches. Since the ham limits the number of sandwiches we can make, the ham is our limiting reagent and the bread is going to be in excess when the ham is consumed by the reaction. Additionally, since the theoretical yield depends on the limiting reactant, we can say that our theoretical yield for the above reaction is six sandwiches. It does not matter that there is enough bread to make 9 sandwiches. Once the ham runs out, information technology is not possible to brand whatsoever more sandwiches. The reaction is consummate at this indicate. Limiting reactant is completely consumed while the excess reactant (bread) is left over. Nosotros could fifty-fifty have this a footstep further and determine the amount of excess reagent left over at the stop of the reaction. In order to perform that adding, use the theoretical yield to calculate the corporeality of excess reactant used in the reaction. Then you lot can subtract the amount of excess reactant used in the chemical equation from the amount you began with.

Example 2

For instance given the counterbalanced reaction [latex]\text{N}_{2}+two\text{H}_{3}\rightarrow2\text{NH}_{3}[/latex]. If yous began with 28 k of N₂ and 2.eight 1000 of H₂. Since information technology is not possible to determine which reactant is the limiting reactant merely from the masses of the reactants, you must first convert the grams to moles using the molecular weights.

Therefore, [latex]\displaystyle28\text{g N}_2=\frac{1\text{ mol Northward}_2}{28.0\text{ g N}_2}=1\text{ mole N}_2\\[/latex]

And [latex]\displaystyle2.8\text{ m H}_2=\frac{1\text{ mol H}_2}{ii.02\text{ g H}_2}=1.iv\text{ mole H}_2\\[/latex]

While there is indeed more H₂ than N_two based on moles of reactants, this is not the final answer! You must convert to the mol of product using the mol to mol ratio.

If N₂ is completely used

1 mole of N_ii [latex]\displaystyle\frac{ii\text{ mol NH}_3}{1\text{ mol North}_2}=\underline{2 \text{mol NH}_{3}}\\[/latex] produced

If H_ii is completely used

1.4 mole of H₂ [latex]\displaystyle\frac{2\text{ mol NH}_3}{three\text{ mol H}_2}=\underline{0.93\text{ mol NH}_{3}}\\[/latex] produced

Thus, since H_ii volition produce less of the product, information technology is the limiting reagent and Due north₂ is the backlog reagent. Here the theoretical yield is 0.93 mol NH₃. Note that for the problems in today's lab we volition then convert the mol of product to grams using its molar mass.

Percent Yield

It is often important to calculate the percentage yield of a reaction. If everything goes according to plan, you volition become exactly 100 percent of the theoretical yield produced in your reaction. However, laboratory errors will oftentimes impact this number. Spills, calculation errors, not drying a product and many other errors bear upon the mass of product obtained. Here the corporeality of product really produced in the laboratory experiment is compared to the corporeality of product that should have been made theoretically. Per centum yield is given by the equation:

[latex]\displaystyle\text{Per centum Yield}=\frac{\text{Actual (Experimental) Yield}}{\text{Theoretical Yield}}\times100\\[/latex]

Guidelines for Limiting Reagent Problems (Calculating Theoretical Yield):

Convert from grams of reactant added to mol using molar mass.
Convert from mol of reactant to mol of product using the coefficients in the balanced equation (mol to mol ratio).
Catechumen from mol of production to mol of reactant using the tooth mass.

Helpful Hints:

These bug must be worked out stoichiometrically.
Yous cannot compare masses of reactants. You MUST convert to mol.
You cannot compare mol of one reactant to another UNLESS you consider the mol to mol ratio. In the reaction between hydrogen and nitrogen, there is technically more mol of hydrogen added to the reaction vessel. Nonetheless once the mol to mol ratio is considered, it becomes apparent that hydrogen is the limiting reactant even though it had more mol.
The molar mass of hydrates MUST include the mass of the water molecules fastened to the ionic chemical compound.

In this experiment, you will predict and observe a limiting reactant during the reaction which involves the reduction of copper (Ii) chloride dihydrate. You will use the single displacement reaction of solid aluminum with aqueous copper (Ii) chloride.

[latex]2\text{ Al(s)}+3\text{ CuCl}_{ii}\cdot2\text{H}_{2}\text{O}(aq)\rightarrow3\text{ Cu (s)}+2\text{ AlCl}_{3}(aq)+half dozen\text{ H}_{2}\text{O(50)}[/latex]

Copper (II) chloride, CuCl₂, turns a light blue in aqueous solution. This is due to the Cu 2+ ion. Aluminum chloride is colorless in aqueous solution. You will be able to monitor the reaction's progress by evaluating the color change occurring in your chalice. The product of solid copper is relevant in many industrial processes. Copper is mankind'due south oldest metallic, dating back more than 10,000 years. The copper (Two) chloride reduction reaction has been used in petroleum industries for sweetening (a refining process used to remove sulfurous gases from natural gas). This procedure has as well been used for etch bath regeneration. In an etch bath, CuCl₂ is used to remove unwanted copper from printed copper coated wiring boards, leaving only copper "wiring."

*Note that the hydrate portion of [latex]\text{CuCl}_{2}\cdot2\text{ H}_{2}\text{O}[/latex] should exist included in the molar mass calculation.