lecturer: amal abu- mostafa. how to recognize which type of chemical reactions. balancing chemical...
TRANSCRIPT
Lecture No. 4Chemical reactions and
chemical equationsLecturer:
Amal Abu- Mostafa
How to recognize which type of chemical
reactions.
Balancing chemical equation.
Calculation based on chemical equation.
Session Objectives:
Look at the reactants and products Element(E), Compound(C)
E + E → C Synthesis C → E + E Decomposition E + C → E + C Single replacement C + C → C + C Double
replacement Acid+Base → H2O + Salt Acid/Base
reaction (Double replacement)
CxHy + O2 → CO2 + H2O (Look at the Products)
Combustion
How to recognize which type of chemical reactions
H2 + O2 → Synthesis H2O → Decomposition AgNO3 + NaCl → Double replacement Zn + H2SO4 → Single replacement HgO → Decomposition NaCl(s) + F2(g) → Single replacement KBr +Cl2 → Single replacement Mg(OH)2 + H2SO4 → Double replacement (Acid/Base) HNO3 + KOH → Double replacement (Acid/Base)
Examples
Zn + O2 ® Synthesis
HgO + Pb® Single replacement HBr + NH4OH ®Double
replacement Cu(OH)2 + KClO3 ® Double
replacement AgBr + Cl2 ®Single replacement
CaPO4 ®Decomposition
Examples
When the coefficients in a chemical equation are correctly given, the numbers of atoms of each element are equal on both sides of the arrow, none are destroyed and none are created. The equation is then said to be balanced.
Important rule: Balance first the atoms for elements that occur in only one substance on each side of the equation.
BALANCING CHEMICAL EQUATION:
As an example, consider the burning of propane gas:
C3H8 + O2 CO2 + H2O (this equation is not balanced).
First C atoms: 1C3H8 + O2 3CO2 + H2O Second H atoms: 1C3H8 + O2 3CO2 + 4 H2O Third O atoms: 1C3H8 + 5 O2 3CO2 + 4 H2O
How to balance this equation:
A) HCl + Na2CO3→NaCl + CO2 + H2O 2HCl + Na2CO3→2NaCl + CO2 + H2O
B) balance C and H then O 1) CH4 + O2 CO2 + H2O CH4 + 2O2 CO2 + 2H2O
2) C8H18 + O2 → CO2 + H2O C8H18 + 25/2 O2 → 8 CO2 + 9 H2O 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
Balance these equations:
H3PO3 → H3PO4 + PH3
Solution: Oxygen occurs in just one of the products (H3PO4). So
it is therefore easiest to balance O atoms first. 4H3PO3 → 3H3PO4 + PH3
This equation is now also balanced in P and H atoms. Thus, the balanced equation is
4H3PO3 → 3H3PO4 + PH3
Balance this equations:
To answer quantitative questions, you must first look at the balanced chemical equation, for example:
N2(g) + 3H2(g) → 2NH3(g)
1molecule N2 + 3 molecules H2 → 2molecules NH3
1 mol N2 + 3 mol H2 → 2 mol NH3
(1×28) g N2 + (3 × 2) g H2 → (2 × 17 )g NH3
CALCULATION BASED ON CHEMICAL EQUATION:
Example 1 : A)How many moles of O2 are needed to burn 1.8 mol of
C2H5OH in the equation? B)And How many moles of CO2 produced when 0.274
mol C2H5OH burned in this example?
C2H5OH + 3O2 → 2CO2 + 3H2O Solution: (A) from the equation: 1mol of C2H5OH needs → 3 mol of O2 to be burned 1.8 mol of C2H5OH needs → ??? Moles of O2
moles of O2= 1.8 × 3 = 5.4 mol 1
CALCULATION BASED ON CHEMICAL EQUATION:
1mol of C2H5OH produces → 2 mol of CO2
0.274 mol of C2H5OH produces → ?? Moles of CO2
Moles of CO2 = 0.274 × 2 = 0.548 mol
1
Solution: (B) from the equation:
A) How many gm of O2 are required to react with 0.3 mol Al in the reaction
4 Al + 3 O2 → 2 Al2O3
Solution: First we find how many moles of O2 are required to react with 0.3 mol Al in the reaction.
4 mol of Al → 3 mol of O2 are required
0.3 mol of Al → ?? mol of O2 are required moles of O2 = 0.3 × 3 = 0.225 mol 4 Mass of O2 = n × M = 0.225× 32= 7.2 g
Example 2:
B) How many gram of Al2O3 could be formed if 12.5 g O2 react completely with Al?
4 Al + 3 O2 → 2 Al2O3
Solution: Moles of O2 = m = 12.5 = 0.391 mol M 32
Example 2:
From the equation: 3 mol of O2 → 2 mol of Al2O3
0.391 mol of O2 → ?? Moles of Al2O3
Moles of Al2O3 = 0.391× 2= 0.26 mol 3 MASS OF Al2O3 (m) = n × M= 0.26 × 102 =26.6 g
Thank you