6.22  This question asks you to convert moles to grams.  To interconvert between moles and grams we use the molar mass of the substance.

(a) Molar mass of CO₂ =  12.01 + 2(16.00) = 44.01 g / mol

2.48 mol CO2	44.01 g CO2	= 109 g CO2
	mol CO2

(b) Molar mass of Au = 196.97 g / mol

0.750 mol Au	196.97 g Au	= 148 g Au
	mol Au

c) Molar mass of CH₄  =  12.01 + 4(1.01) = 16.05 g / mol

5.40 x 10-2 mol CH4	16.05 g CH4	= 0.864 g CH4
	mol CH4

6.24 This question asks you to convert grams to moles.  To interconvert between moles and grams we use the molar mass of the substance.  Note that we divide by the molar mass whereas in problem 6.22 we multiplied by the molar mass (just make the units cancel)

a) Molar mass of S = 42.06 g / mol

5.23 g S	mol S	= 0.166 mol CO2
	42.06 g S

(b) Molar mass K₂O = 2(39.10) + 16.00 = 94.20 g / mol

1.45 g K2O	mol K2O	= 0.0154 mol K2O
	94.20 g K2O

c) Molar mass of C₂H₆  =  2(12.01) + 6(1.01) = 30.08 g / mol

0.0500 g C2H6	mol C2H6	= 0.00167 mol C2H6
	30.08 g C2H6

6.28  Rewrite these word equations as balanced chemical equations: 

(a) 2Na_(s) + Cl_2(g) -> 2NaCl_(s)

This the classic reaction of a metal and non-metal reacting to form an ionic compound sodium chloride.  Generally, when we write out the names of compounds we never write down the name of just a single ion.  Hence, if we see sodium we know it is elemental sodium Na and not sodium ion Na⁺.  You need to remember which elements are diatomic (i.e. H₂, see lecture 9) to recognize that chlorine is written as Cl₂(g).

(b) S_(s) + O_2(g) -> SO₂

Despite the anion sounding name of oxide, sulfur dioxide is covalent compounds made from the two non-metals sulfur and oxygen.  Remember ionic compounds generally form from a metal and non-metal.  In naming covalent compounds of oxygen the mono, di, tri prefixes are often used to indicate the number of oxygens (1,2,3, respectively)

(c) CaCl₂ + Na₂SO₄ -> 2 NaCl + CaSO₄

To write the formula for the ionic compounds, it is necessary to know the ions formed by a variety of elements as well as some common polyatomic ions.  For example to know that calcium chloride is CaCl₂, you need to know that calcium generally forms a 2+ cation and chlorine generally a 1- anion.   Remember,  for elemental anions and cations, the charge is often rationalized by considering the number of electrons gained or lost to attain Nobel gas configuration.

(d) CaCO₃ -> CaO + CO₂

6.30 In doing stoichiometry questions, the key is using the coefficients of the chemical reaction to relate moles of one substance to moles of another.

(a)

Given: moles O₂

Asked to find moles CO₂

This problem requires us to relate moles of CO₂  with moles of O₂.  We will use the coefficients of the balanced chemical equation given to generate a pair of "conversion factors" that we can use to interrelate these two.

We use the first of these two so that we get the proper cancellation of units.

(b) Given moles C₂H₂

Asked to find moles O₂

Similar to (a) but we use  a different conversion factors

(c) Given: mass of CO₂

Asked to find grams of H₂O

We know are asked to find amounts in grams.  When using the coefficients of the chemical reaction, we need to work in moles so we are going to need to interconvert between moles and grams using molecular masses.

Molar mass of CO₂ = 12.01 + 2(16.00) = 44.01 g / mol

Molar mass of H₂O = 2(1.01) + 16.00 = 18.02 g / mol

(d) Given 6.0 g water

Asked to find grams C₂H₂(g)

Molar mass H₂O calculated in (c)

Molar mass of C₂H₂ = 2(12.01) + 2(1.01) = 26.04 g / mol

6.36  A product of the reaction is 192 kcal of energy.  This means the reaction gives of 192 kcal of energy as goes from reactants to products.  Giving off energy corresponds to a negative DG.  The free energy of the products is lower than the free energy of the reactants.  We can represent this with a free energy diagram as shown below:

6.56  The equilibrium reaction given is:

PCl₅(g) + heat PCl₃(g) + Cl₂(g)

This problem tests your understanding of LeChatelier's principle.  In an equilibrium reaction, if we add more reactants, the reaction counteracts this "stress" by converting some of the added reactants to products.  Conversely, if we add more products, the reaction conteracts this "stress" by converting some of the added products to reactants.  Note, that heat can be treated just like a product of reactant.

(a) The addition of Cl₂ will result in the formation of more reactants:  PCl₅(g) + heat.

(b) The removal of heat will pull the reaction to form more reactants: PCl₅(g) + heat

(c) The removal of PCl₃ (g) will pull the reaction to continue to form more products PCl₃(g) + Cl₂(g)

6.60

The rate of a chemical reaction is affected by the activation energy, temperature, concentration, surface area for solids involved in reactions, and the presence of a catalysts.  The activation energy is a characteristic of a particular reaction and is influenced by the addition of a catalyst. 

(a) Combustion is the reaction of a carbon compound with oxygen to form mainly carbon dioxide and water.   Fanning the fire replenishes the supply of oxygen (increases its concentration) thereby driving the reaction faster (bigger flame!)

(b) Dust has a large surface area and consequently its combustion can occur so fast that it is explosive.

(c) The manganese dioxide is not in itself changed by it accelerates the decomposition of hydrogen peroxide.   Consequently, it is a catalyst.

6.64

The equilibrium constant for an equilibrium chemical reaction describes the relative amounts of products and reactants present at equilibrium.  For a general reaction:

aA + bB cC + dD

the equilibrium constant is given by

where the brackets [ ] indicate concentration.  For the reaction given in the problem, the equilibrium constant is

(a) K_eq < 1 x 10^-3 implies that the [B] must be very small relative to [Y].  A small [B] and a big [Y] means that the ratio [B] / [Y] will be small.  Hence the equilibrium position is where there is a relatively large amount of [Y] and the solution will be yellow.

(b) K_eq = 1.  For K_eq =1, the [B] = [Y] so that the ratio [B] / [Y] = K_eq = 1.  Hence, here there is an equal amount of [B] and [Y].  Since yellow and blue make green, this solution will be green in color. 

(c) K_eq < 1 x 10³ implies that the [B] must be very large relative to [Y].  A large [B] and a small [Y] means that the ratio [B] / [Y] will be large.  Hence the equilibrium position is where there is a relatively large amount of [B] and the solution will be blue.