Understanding Points |
Reactivity 2.1.1—Chemical equations show the ratio of reactants and products in a reaction. |
Reactivity 2.1.2—The mole ratio of an equation can be used to determine:
• the masses and/or volumes of reactants and products
• the concentrations of reactants and products for reactions occurring in solution. |
Reactivity 2.1.3—The limiting reactant determines the theoretical yield. |
Reactivity 2.1.4—The percentage yield is calculated from the ratio of experimental yield to
theoretical yield. |
Reactivity 2.1.5—The atom economy is a measure of efficiency in green chemistry. |
Balancing Chemical Equations
•
Law of conservation of mass: mass cannot be created nor destroyed in chemical reactions
◦
Number of atoms in reactant = number of atoms in product
◦
Reactants → Products
•
Balancing Equations Guide:
◦
Count the number of atoms in both the reactant and product side
◦
Place coefficients as needed such that the number of atoms on each side are the same– do NOT change the subscript in the formulas
▪
Changing the coefficient changes the chemical substance (compound or element) involved in the reaction
•
Examples:
Mg + AgCl → MgCl2 + Ag
(Metal displacement reaction)
Reactants | Products |
Mg + AgCl → MgCl2 + Ag | |
1 x Mg
1 x Ag
1 x Cl | 1 x Mg
1 x Ag
2 x Cl |
Since Cl is not balanced, 2 can be added as a coefficient in front of AgCl:
Mg + 2AgCl → MgCl2 + Ag | |
1 x Mg
2 x Ag
2 x Cl | 1 x Mg
1 x Ag
2 x Cl |
Ag still not balanced– 2 can be added as a coefficient in front of Ag:
Mg + 2AgCl → MgCl2 + 2Ag | |
1 x Mg
2 x Ag
2 x Cl
= to product side | 1 x Mg
2 x Ag
2 x Cl
= to reactant side |
C6H12O6 + O2 → CO2 + H2O
(Combustion of glucose/cellular respiration)
Products | Reactants |
C6H12O6 + O2 → CO2 + H2O | |
6 x C
8 x O
12 x H | 1 x C
3 x O
2 x H |
Since C is not balanced, 6 can be added as a coefficient in front of CO2:
C6H12O6 + O2 → 6CO2 + H2O | |
6 x C
8 x O
12 x H | 6 x C
13 x O
2 x H |
H is also not balanced, so 6 can be added as a coefficient in front of H2O:
C6H12O6 + O2 → 6CO2 + 6H2O | |
6 x C
8 x O
12 x H | 6 x C
18 x O
12 x H |
C6H12O6 + 6O2 → 6CO2 + 6H2O | |
6 x C
18 x O
12 x H
=product side | 6 x C
18 x O
12 x H
=reactant side |
Limiting Reagent: the reactant that is fully used up in the reaction and leaves the remaining reactants not fully used
•
All the reactants that are unused by the end of the reaction are in excess; i.e. there is too much of the reactants in excess
Example:If there are 2 moles of NH3 and 1 mole of O2 in the below reaction, identify which reactant is limiting and which is in excess.
4NH3(g) + 5O2 (g) → 4NO (g) + 6H2O (l)
For every 4 NH3 molecules, 5 O2 molecules react.
NH3O2=45 and NH3 = 2 moles; O2 = 1 mole
NH3=45(O2) meaning that for 1 mol of NH3, 0.8 mol of O2 is required;
Likewise, for 2 mol of NH3, 1.6 mol of O2 is required
But since there is only 1 mol of O2 in the reaction and 2 mol of NH3, NH3 is in excess and O2 is the limiting reagent.
Yield
1.
Percentage yield = experimental yield theoretical yield 100
2.
Experimental Yield: The actual measured amount of product from the experiment
3.
Theoretical Yield: the calculated expected yield using the amount of reactants used
•
The number of moles of product formed from reacting all the limiting reagent
•
How to calculate the theoretical yield:
For any given balanced reaction:
aL + bE → cP
where L: limiting reagent, E: reagent in excess, P: product
Reactant | Product | |
Mass / conc. /vol. | ||
Mole | ||
Ratio |
1.
Determine the number of moles of limiting reagent (n= mass/molar mass)
2.
Determine the ratio of limiting reagent to product
a.
Ratio of L to P: ac
3.
Multiply the number of moles of limiting reagent with the ratio of limiting reagent to product (= number of moles of product)
a.
n(L) × ac = n(P)
4.
Multiply the result of step 3 with the molecular weight of the product
a.
n(P) × Mr(P) = n(L) × ac x Mr(P) = theoretical yield (mass)