Section 13.1Balancing Equations
1. (Exercises 1 and 2) Identify each of the following as a physical or chemical change:

(a) mixing concentrated orange juice with water;

(b) neutralizing spilled acid by mixing it with baking soda;

(c) adding salt to your french-fries at McDonald's.

The above interactions are (a) physical, (b) chemical, and (c) physical.

2. (Exercises 5 and 6) Identify which of the following reactions is a combination reaction, which is a decomposition reaction, and which is a hydrocarbon combustion reaction.

(a) CH₄ + 2O₂ --> CO₂ + 2H₂O

(b) 2 Mg + O₂ --> 2 MgO

(c) 2 H₂O --> 2H₂ + O₂

Answers: (a) hydrocarbon combustion, (b) combination, and (c) decomposition.

3. (Exercises 7 and 8) Write and balance the equations for the following reactions.

(a) Potassium chlorate (KClO₃) decomposes when heated to form KCl and O₂.

This would be a very simple equation to balance if we could just place a 3 as the subscript of an oxygen atom, O, on the right side, but since oxygen forms a diatomic molecule we must balance this equation by using a coefficient 3 in front of the O₂ and placing 2's in front of both other formulas, as shown below.

2 KClO₃ --> 2 KCl + 3 O₂

(b) Iron and water react to form Fe₃O₄ and free hydrogen.

The basic equation can be written as follows:

Fe + H₂O --> Fe₃O₄ + H₂

The balanced form of this equation requires 3 iron atoms and 4 water molecules to form one Fe₃O₄ molecule plus 4 hydrogen molecules.

3 Fe + 4 H₂O --> Fe₃O₄ + 4 H₂

Section 13.3Acids and Bases
4. (Exercises 11 and 12) Complete and balance the following acid-base reactions.

(a) HCl + KOH -->

(b) H₃PO₄ + Ba(OH)₂-->

In each of these reactions the acid and the base react to produce water and a salt, so first put H₂O on the product side and then write the formula for the salt that is formed. Then balance the equation.
(a) In this simple reaction the equation needs no coefficients in front of any of the compounds and the basic equation provides the final solution. A check of the equation shows that there is an equal number of atoms of each element on both sides of the equation shown below, so the equation is balanced as written.

HCl + KOH --> H₂O + KCl

(b) The salt formed is Ba₃(PO₄)₂, which requires a 2 in front of the H₃PO₄ and a 3 in front of the Ba(OH)₂. To make the final balanced equation, put a 6 in front of the H₂O.

2 H₃PO₄ + 3 Ba(OH)₂ --> 6 H₂O + Ba₃(PO₄)₂

Note that there are 12 hydrogen atoms and 6 oxygen atoms on each side, and 3 Ba atoms and 2 polyatomic ions of (PO₄) on each side of the balanced equation. Considering the PO₄^3- polyatomic ion as a unit makes the balancing process easier.

Section 13.4Single-Replacement Reactions
5. (Exercises 15 and 16) Refer to the activity series (Table 13.4 in the textbook) and predict in each case whether the single-replacement reaction shown will, or will not, actually occur.

(a) 2 Al + 3 FeSO₄(aq) --> 3 Fe + Al₂(SO₄)₃(aq)

(b) 2 Ag + FeSO₄(aq) --> Fe + Ag₂SO₄(aq)

(a) In this reaction Al is more active (higher in the table) than Fe, so the reaction will occur as written.
(b) In this reaction Ag is less active (lower in the table) than Fe, so the reaction will not occur.

Section 13.5Avogadro's Number
6. (Exercises 19 and 20) Calculate the number of molecules in 3.00 moles of (a) water, and (b) sulfuric acid.

(a) Each mole of any substance contains Avogadro's number of atoms or molecules, so the 3.00-mole sample of water will contain:

# molecules of water = 3.00 moles (6.02 x 10²³ molecules/mole)

= 18.1 x 10²³ molecules

(b) Because the question specifies the same number of moles of sulfuric acid, 3.0 moles, the number of molecules of acid will be exactly the same.

# molecules of acid = 3.00 moles (6.02 x 10²³ molecules/mole)

= 18.1 x 10²³ molecules

7. (Exercises 21 and 22) What will be the mass of a 3.00 mole sample of (a) water and (b) sulfuric acid?

(a) The chemical formula for water is H₂O, so its formula mass will be 2 x 1.01 u for the two hydrogen atoms, plus 1 x 16.0 u for the one oxygen atom. This gives a total formula mass of 18.0 u. The molar mass of water is, therefore, 18.0 g/mol. Since there are 3.00 moles present in this sample, the mass will be:

mass = 3.00 moles (18.0 g/mole) = 54.0 g

(b) The chemical formula for sulfuric acid is H₂SO₄, so its formula mass will be 2 x 1.01 u, plus 1 x 32.1 u, plus 4 x 16.0 u = 98.1 u. This means that the molar mass of sulfuric acid will be 98.1 g/mol. Since there are 3.00 moles present in this sample, the mass will be:

mass = 3.00 moles (98.1 g/mole) = 294 g

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