# Percent Yield Lab. This experiment has proven that KI is the limiting reagent in this chemical reaction.

by honeybooboo (student)

Percent Yield Lab

Introduction

When a chemical reaction occurs and two reactants mix, we should in theory be able to determine the amount of each product formed using stoichiometric calculations. The maximum value of the final product is limited to the limiting reagent not the reagent in excess. From determining the limiting reagent we are able to calculate the theoretical mass of the product formed.

First step to find the percentage yield is to balance. The number of atoms in the reactants must be equal to the atoms in the products. This is due to the law of conservation of mass. Matter can’t be destroyed or created but transferred so therefore the chemical equation has to be balance so that the number of atoms on both side and the exact ratio of how the reactants is displayed.

Second step is to find the limiting reagent. Limiting reagent is the reactant that is depended upon to determine how much of product is made. In order to find the limiting reagent the equation moles= mass/molar mass. Now that the moles are figured out, multiply the number of moles of the reactants by the ratio of the reactant and the product. This calculation is performed twice once for the first reactant and one for the second. Then the Theoretical Yield is calculated, it is how much product will be synthesized with the reactants. Multiply the lowest number of moles (limiting reagent’s mole) by the molar mass of the product. This will give you the mass of the product.

In order to find the percent yield, a ratio between the actual yield and the theoretical yield is used. This indicates the percent of how much of the theoretical yield was obtained in the experiment. This equation can be used: Percentage yield= mass of actual yield/ mass of theoretical yield * 100%.

Purpose:

To quantitatively and qualitatively determine the limiting reagent in the chemical reaction. To also compare the theoretical and actual values in order to find the percentage yield of the reaction.

Hypothesis

Pb (NO3)2(aq)           +               2KI (aq)                           PbI2 (s) + 2KNO3

Test #1 (Pb (NO3)2)

xPbI20.00906mol=1/1

X= 0.00906mol

Test #2 (KI)

xPbI20.0120mol=1/2

X= 0.00602mol

Theoretical Yield

0.00602*(461) = 2.78 g

The limiting reagent should be KI. KI has the least amount of moles therefore you do not have ...