LAB 6D

Determine the Limiting Reactant and Percent Yield in a Precipitation Reaction

Objectives

• to observe the reaction between solutions of sodium carbonate and calcium chloride
• to determine which of the reactant is the limiting reactant and which is the excess reactant
• to determine the theoretical mass if precipitate that should form
• to compare the actual mass with the theoretical mass of precipitate and calculate the percent yield

After we put two kinds 25ml of solution: Na2CO3 and CaCl2, the precipitate will form.



Set up the filtering apparatus like the picture show above. And pure the mixture slowly and carefully into the filter funnel. It takes some time to complete the filtering process.Then put the filter paper away and dry it.

After the filter paper is completely dry, we can weigh the mass of filter paper to calculate the mass of precipitate.

Accoding to the molarity of the solution we used for this lab, we can calculate the mass of Na2CO3 and CaCl2. According to chemical equation, we can determine which one is the limiting reactant and which one is excess reactant.

Then we can calculate the mass of CaCO2 we expect to get.
Use the formula % yield= mass actual produced/mass expect produced*100%, we can calculate the % yield of this lab.

Percent Yield & Purity

PERCENT YIELD


Reactions rarely produce the predicted amount of product from the masses of reactants in the reaction .



The percent yield is defined as




 

The predicted yield is determined by the masses used in a reaction and the mole ratios in the balanced equation. This predicted yield is the "ideal". It is not always possible to get this amount of product.

1. Balancing The Chemical Equation:

The first step in finding theoretical and percentage yield is to balance the relevant chemical equation.

2. Finding The Limiting Reagent:

• this is the reactant which the product yield depends on, as it is not in excess.

• To determine which reactant is the limiting reagent

1(a). Divide the mass (in grams) of the reactant by its molecular weight (in g/mol) OR
1(b). Multiply the amount used (in mL) by its density, then divide by its molar mass
2. Multiply the mass (your answer from steps 1(a) or 1(b)) by the number of moles of the reactant used in the reaction.

3. Percent Yield

The percentage yield is the ratio between the actual yield and the theoretical yield multiplied by 100%. It indicates the percent of theoretical yield that was obtained from the final product in an experiment. 

Finding Percent Purity

When we make something in a chemical reaction, and separate it from the final mixture, it will still have small amounts of other substances mixed with it. It will impure.

The formula for percent purity is:

Example:

The aspirin from the above experiment was not pure. 121.2 g of solid was obtained, but analysis showed that only 109.2g of it was aspirin. Calculate the percent purity of the product.

Solution:

Percent purity = 109.2 ÷ 121.2 × 100% = 90.0%

EXCESS & LIMITING REACTANT

Excess and Limiting reactant

2H₂ + O₂ 

→

 2H₂O

If 4 grams of hydrogen reacts with 16 grams of oxygen what mass of water is formed?
Step 1 Convert to mole all the reactants
Moles of hydrogen = 4grams/2 atomic mass units = 2mole
Moles of Oxygen gas = 16 grams/ 32 atomic mass units = 0.5mole

Step 2 Identify the reactant that is the LIMITING reactant. (In other words the reactant that will stop the reaction and the formation of product)
Take hydrogen and ask the following questions.How much do we have? = 2moles How much oxygen do we need to react with this amount of hydrogen? = 1/2 X 2moles = 1mole of oxygen.
 we have 0.5 mole of oxygen, therefore oxygen is our limiting reactant.

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O + energy
 
What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?
 
When approaching this problem, you should be able to see that for every 1 mole of glucose (C6H12O6), you need 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.
 
Step 1: Determine the balanced chemical equation for the chemical reaction.
 
Already given!
 
Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).
 
25g x (1 mol/180.06g) = 0.1388 mol  C6H12O6
 
40g x (1 mol/32g) = 1.25 mol O2                                                                                                                                               
 
Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio.
 
a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25 x (1/6) or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.
 
     
 
1.25 mol O2 x (1 mol C6H12O6)/(6 mol O2) = 0.208 mol C6H12O6
 
b. If all of the 0.1388 moles of glucose were to be used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Since there is an excess of oxygen, the glucose amount will be used to calculate the amount of the products in the reaction.
 
0.1388 mol C6H12O6 x (6 mol O2)/(1 mol C6H12O6) = 0.8328 mol O2
 
If more than 6 moles of O2 is available per mole of C6H12O6, the oxygen is in excess and the glucose is the limiting reactant. If less than 6 mole of oxygen is available per mole of glucose, the oxygen is the limiting reactant. Our ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 . 
 

Therefore the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6)
 
Which gives us a 4.004 ratio of O2 to C6H12O6.
 
Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.
 
For carbon dioxide produced: 0.1388 moles glucose x (6/1) = 0.8328 moles carbon dioxide.
 
Step 5: If necessary, calculate how much is left in excess.
 
1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over