Lecturer: Amal Abu- Mostafa. How to recognize which type of chemical reactions. Balancing chemical equation. Calculation based on chemical equation.

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Lecturer: Amal Abu- Mostafa

How to recognize which type of chemical reactions. Balancing chemical equation. Calculation based on chemical equation.

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 H 2 O + Salt Acid/Base reaction (Double replacement) C x H y + O 2 CO 2 + H 2 O (Look at the Products) Combustion

H 2 + O 2 Synthesis H 2 O Decomposition AgNO 3 + NaCl Double replacement Zn + H 2 SO 4 Single replacement HgO Decomposition NaCl (s) + F 2(g) Single replacement KBr +Cl 2 Single replacement Mg(OH) 2 + H 2 SO 4 Double replacement (Acid/Base) HNO 3 + KOH Double replacement (Acid/Base)

Zn + O 2 Synthesis HgO + Pb Single replacement HBr + NH 4 OH Double replacement Cu(OH) 2 + KClO 3 Double replacement AgBr + Cl 2 Single replacement CaPO 4 Decomposition

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.

As an example, consider the burning of propane gas: C 3 H 8 + O 2 CO 2 + H 2 O (this equation is not balanced). First C atoms: 1C 3 H 8 + O 2 3CO 2 + H 2 O Second H atoms: 1C 3 H 8 + O 2 3CO 2 + 4 H 2 O Third O atoms: 1C 3 H 8 + 5 O 2 3CO 2 + 4 H 2 O

A) HCl + Na 2 CO 3 NaCl + CO 2 + H 2 O 2HCl + Na 2 CO 3 2NaCl + CO 2 + H 2 O B) balance C and H then O 1) CH 4 + O 2 CO 2 + H 2 O CH 4 + 2O 2 CO 2 + 2H 2 O 2) C 8 H 18 + O 2 CO 2 + H 2 O C 8 H 18 + 25/2 O 2 8 CO 2 + 9 H 2 O 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O

H 3 PO 3 H 3 PO 4 + PH 3 Solution: Oxygen occurs in just one of the products (H 3 PO 4 ). So it is therefore easiest to balance O atoms first. 4H 3 PO 3 3H 3 PO 4 + PH 3 This equation is now also balanced in P and H atoms. Thus, the balanced equation is 4H 3 PO 3 3H 3 PO 4 + PH 3

To answer quantitative questions, you must first look at the balanced chemical equation, for example: N 2 (g) + 3H 2 (g) 2NH 3 (g) 1molecule N 2 + 3 molecules H 2 2molecules NH 3 1 mol N 2 + 3 mol H 2 2 mol NH 3 (1×28) g N 2 + (3 × 2) g H 2 (2 × 17 )g NH 3

Example 1 : A)How many moles of O 2 are needed to burn 1.8 mol of C 2 H 5 OH in the equation? B)And How many moles of CO 2 produced when 0.274 mol C 2 H 5 OH burned in this example? C 2 H 5 OH + 3O 2 2CO 2 + 3 H 2 O Solution: (A) from the equation: 1mol of C 2 H 5 OH needs 3 mol of O 2 to be burned 1.8 mol of C 2 H 5 OH needs ??? Moles of O 2 moles of O 2 = 1.8 × 3 = 5.4 mol 1

1mol of C 2 H 5 OH produces 2 mol of CO 2 0.274 mol of C 2 H 5 OH produces ?? Moles of CO 2 Moles of CO 2 = 0.274 × 2 = 0.548 mol 1

A) How many gm of O 2 are required to react with 0.3 mol Al in the reaction 4 Al + 3 O 2 2 Al 2 O 3 Solution: First we find how many moles of O 2 are required to react with 0.3 mol Al in the reaction. 4 mol of Al 3 mol of O 2 are required 0.3 mol of Al ?? mol of O 2 are required moles of O 2 = 0.3 × 3 = 0.225 mol 4 Mass of O 2 = n × M = 0.225× 32= 7.2 g

B) How many gram of Al 2 O 3 could be formed if 12.5 g O 2 react completely with Al? 4 Al + 3 O 2 2 Al 2 O 3 Solution: Moles of O 2 = m = 12.5 = 0.391 mol M 32

From the equation: 3 mol of O 2 2 mol of Al 2 O 3 0.391 mol of O 2 ?? Moles of Al 2 O 3 Moles of Al 2 O 3 = 0.391× 2= 0.26 mol 3 MASS OF Al 2 O 3 (m) = n × M= 0.26 × 102 =26.6 g

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