Excess & Limiting Reactants persent yield

A.Excess Quantity
A balanced equation describes what should happen in a chemical reaction.

However, the conditions necessary for the reaction to take place may not be present. (ie. pressure, temperature, concentration etc.)

Sometimes it is necessary to add more of one reactant than the equation predicts because it is not possible for every atom/molecule of the reactants to come together.
Ex. 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 reactant.

B.Excess quantities in chemical reactions.
Ex1. How many grams of OCl2 will be formed when 44.0 g of O2 reactant with 97.0g of Cl2

(*General Process: Convert both reactants to the desired product & the smaller amount of product will actually be produced.)

Step1: Balanced equation.
O2(g) +2 Cl2(g) → 2OCl2(g) 

Step2: Convert Cl2 to OCl2
97.0 g ×1mol Cl2/71g × 2mol OCl2/2mol Cl2 × 87.0g/1mol OCl = 118.86 g OCl2

Step3: Convert O2 to OCl2
44.0g × 1molO2/32g ×2mol OCl2/1mol O2 × 87.0g/1mol OCl = 239.25g OCl2
Because 97.0g of Cl2 reacts to produce the smaller amount of product. Cl2 is limiting quantity and O2 is excess quantity. 119g is the mass of product.

Ex2. 
41g O2 reacts of 164g of Cl2, Which is EXCESS quantity? Which is Limiting quantity? How many grams of the excess quantity will be used?
Step1:
O2(g) +2 Cl2(g) → 2OCl2(g) 
41.0g       164g
(*General process: Convert ONE of products to the other reactant to see which excess and which is limiting. Determine which is left over and by how much. It may involve another conversation)
Step2:
To find which reatant is in excess calculate how many grams of Cl2 gas would be required to react with 41.0g of O2.
41g O2 × 1mol O2/32.0gO2 × 2mol Cl2/1mol H2 × 71.0g Cl2/1mol Cl2= 181.93g=182g Cl2

Since there are only 164g of Cl2 gas, not all O2 can react. Thus Cl2 is limiting quantity and O2 is excess quatity.

Step3: Convert Cl2 to O2 to see how much O2 would be needed to react with 164g of Cl2.
164g Cl2 × 1mol Cl2/71.0g Cl2 × 1mol O2/2mol Cl2 × 32.0g/1mol O2 = 36.958=37.0 g O2

37.0g of O2 would react with 164g of Cl2.

STOICHIOMETRY (2) !!

Example 1 :
a) How many grams of Fe(Ⅱ) will be needed to react with 3.20 mol of HCl?

Step1: Balance the equation.
1Fe + 2HCl → 1FeCl2 + 1H2

Step2: Make a "ROAD MAP"

Step 3: Do calculation

3.2 mol HCl × 1mol Fe/ 2mol HCl × 56g Fe/ 1mol Fe = 90g Fe

b) How many grams of Fe will be needed to produce 10.0 g H2

10.0 g × 1mol/2.0 g H2 × 1mol Fe/ 1mol H2 × 56 g Fe/1 mol Fe =280 g Fe

Note: grams (X) ↔ mol(x) ↔ mol(y) ↔ grams(y)

Example 2 :

a) Using the following equation : 2 NaOH + 1H2SO4 → 2H2O + 1Na2SO4

How many grams of sodium sulphate will be formed if you start with 200 grams of sodium hydroxide?

200 g NaOH × 1mol NaOH/ 40 g × 1mol Na2SO4/2 mol NaOH × 355.3g Na2SO4/ 1mol = 888.25 g

                                                                                                                                            = 900 g

STOICHIOMETRY (1) !!

What is stoichiometry?
stochio=Greek for element
metry=measurement

Stoichiometry is a branch of chemistry that deals with the quantitative analysis of chemical reactions and is about measuring the amounts of the elements and compounds involved in a reaction.

It is the study of the relationship between the amount of reactants uesd in a chemical reaction and the amounts of products produced by the reaction.

Consider the chemical reaction:
4NH3+5O2→6H2O=4NO
with stoichiometry, we find out:
4:5:6:4
do more than just multiply atoms
4:5:6:4
are what we call a mole ratio.

Balanced chemical equations are required.
It tells us the ratio of the molecules or moles of the substances in a chemical reaction.

ex
For the equation: Zn(s)+2HCl(aq)→ZnCl2(aq)+H2(g)
a) How many atomes of Zn are needed to produce 1 molecule of Hydrogen?
1 molecule H×1atom Zn/1molecule H=1atom Zn

b)How many moles of HCl are needed to produce 0.452 moles of Zinc chloride?
0.452mole ZnCl2×2mole HCl/1mole ZnCl2=0.904mole HCl

c)How many grams of Zn will react with 1.05 moles of HCl?
1.05mole HCl×1mole Zn/2mole HCl=0.525mole Zn  0.525mole Zn×65.4g/1mole=0.038mole H