Determing the Limiting Reactant and Percent Yield in a Precipitation Reaction

Lab 6D:

 Determing the Limiting Reactant and Percent Yield in a Precipitation Reaction

The EQUIPMENT we used:

Centigram balance               

2 graduated cylinders (25mL)         

2 beaker (250mL)

1 Wash bottle                 

1 filtering apparatus (ring with stand, Erlenmeyer flask (250mL) +funnel)

1 Filter paper           

lab apron               

safety goggles

Things we need to KNOW before the experiment:

Na2CO3 (aq)+ CaCl2 (aq) → 2NaCl (aq) + CaCO3 (s)

0.70M sodium carbonate solution             

0.50M calcium chloride solution

Objective: 

 To observe the reaction between solutions of sodium carbonate and calcium chloride

 To determine which of the reactants is the limiting reactant  and which is the excess reactant 

 To determine the theoretical mass of precipitate that should form

To compare the actual mass with the theoretical mass of precipitate and calculate the percent yield

Molarity and Storchiometry

Molarity and Storchiometry

Eg. How many grams of Ag will be formed when 100ml of 0.500M AgNO3 is reacted with sufficient Mg?

                 Mg + 2AgNO3 →2Ag + Mg (NO3)2

- first, we can calculate how many moles is contained in the reaction

 100ml=0.100L*0.500M=0.05moles of AgNO3

- Then we can convert moles AgNO3 to moles of Ag and converts to grams Ag

0.05moles AgNO3 *(2mole Ag/ 2 mole AgNO3) * (107.9g/1 mole Ag) =5.40g of Ag

Eg2: How many mL of 0.300M AgNO3 are required t oreact with 3.00 grams of Mg?

- Equation:     Mg + 2AgNO3 → 2Ag + Mg(NO3)2

- Then we can covert grams Mg to moles Mg. Moles Mg to moles AgNO3 and to ML

3.00g Mg * (1 mole Mg/24.3g) * (2 mole AgNO3 / 1 mole Mg) =0.247 moles of AgNO3

 0.247Moles AgNO3 / 0.300M = 0.823L of AgNO3 = 823mL

Gas Stoichiometry

Eg. Water decomposes into H2 and O2, how many grams of H2O are needed to produce 5.0L of O2 measured at STP?

                              2H2O → 2H2 + O2

5.0L * (1 mole O2 / 22.4L) *(2 moles H2O / 1 mole O2)* (18g/ 1 mole H2O) =8.0g of H2O

-Here we convert L to moles(in STP) and then converts mole of O2 to mole of H2O and to grams of H2O

Mar 11--Excess and Limiting Reactants Percent Yield

Imagine yourself making cheese sandwiches.

Excess Quantity: ie. pressure, temperature, concentration, etc. of a chemical reaction which don’t show in the balanced equation.

Since it is not possible for every atom or molecules of the reactants to come together in a chemical reaction, it is necessary sometimes to add more of one reactant than the equation predicts.

ie. One reactant is the Excess Quantity and some of it will be left over. The second reactant is used up completely, and is the Limiting Factor.

In Chemical Reactions, we calculate the amount of product formed when one of the reactants is excess.

Ex. How many grams of S is produced when 15.6 g of KMnO₄, 7.95 g of H₂S, and 15.3 g of H₂SO₄ are reacted?

Step #1--Balance the equation.

2 KMnO₄ + 5 H₂S + 3 H₂SO₄ --> K₂SO₄ + 2 MnSO₄ + 8 H₂O + 5 S

Step #2--Convert all reactants to the desired product and the smallest amount of product will actually be produced.

15.6 g KMnO₄ × (1 mol KMnO₄ / 158.0 g) ×(5 mol S / 2 mol KMnO₄) × (32.1g / 1 mol S) = 7.92 g S

7.95 g H₂S × (1 mol H₂S / 34.1 g) × (5 mol S / 5 mol H₂S) × (32.1g / 1 mol S) = 7.48 g S

15.3 g of H₂SO₄ × (1 mol H₂SO₄ / 98.1 g) × (5 mol S / 3 mol H₂SO₄) × (32.1g / 1 mol S) = 8.34 g S

∴7.48 g S is produced.

STILL GOT QUESTIONS???

Ask Rosengarten!!!!!!!!!!!!!!!!!!!!!!!! :)

March 4--Stoichimetry Calculation

Part 1: Moles and Mass

For example,


Given equation:

I2 + F2  produced   IF5 + I4F2

How many grams of I4F2 are produced by 5.40 mol of F2?

Step1 : Balance the equation

3 I2 + 6 F2 produced 2 IF5 + I4F2

Step 2:Make a map

5.40 mol of F2 ----- to ------ ? grams of I4F2

                  mole I4F2 

Step 3:Do calculation

5.40 mol F2* (1 mole I4F2/ 1 mole F2)* (545.6g of I4F2/1 mole of I4F2)

= 2950g of I4F2

March 1--Stoichiometry

What does STOICHIOMETRY mean?

（”stochio”= Greek for element;  “metry”= measurement）

Stoichiometry = measuring the amounts of elements and compounds involved in a  chemical reaction = relationship between the amount of reactants and products

Consider the equation: 2Mg + O₂ --> 2MgO

Mole Ratio of the above equation = 2: 1: 2

(2 moles of Mg react with 1 mole of O₂ to produce 2 moles of MgO.)

- It tells us the ration of the molecules in an equation

- The ratio can also be used as conversion factors

(What you need over what you have OR Where you’re going over where you start.)

Conversion

Ex. Write the molar ratios for N₂ to H₂ and NH₃ to H₂.

N₂ + 3 H₂ --> 2 NH₃

- N₂ to H₂: 3/1

- NH₃ to H₂: 2/3

Ex. How many moles of O₂ will be formed when 5.1 moles of H₂O is decomposed?

2 H₂O --> 2 H₂ + O₂

- 5.1 mole H₂O×1mole O₂ / 2 moles H₂O = 2.6 moles O₂