__Topics:__

Previous Topic (The Mole)

Chemical reactions also involve substances in the gas phase, therefore to determine the number of moles of a gaseous compound the ideal gas equation is used

PV = nRT where P → pressure in kPa

V → volume in litres (L)

n → number of moles

T → temperature in Kelvin (K)

R
→ universal gas constant;
8.314 kPa L
mol^{1} K^{1}

-the ideal gas law can be used to determine any one of the variables from the knowledge of the other three

Example:

A bulb with a volume of 225
mL contains 0.580 g of an unknown gaseous compound.

The pressure is measured as 145.60 kPa at a temperature of 25 ^{o}C.
What is the molar mass of the compound?

__Solution:__

* *

*Convert 225 mL to L; 225
mL = 0.225 L*

*Convert 25 ^{o}C
to Kelvin; 25 ^{o}C = 273 + 25= 298 K*

* *

*Determine the number of
moles of gas under the above conditions*

* *

*n = PV = 145.0
kPa *

* RT 298 K *x*
8.314 kPaLmol ^{1}K^{1}*

* = 1.32 *x*
10 ^{2}mol*

* *

*1.32 *x* 10 ^{2}mol
= 0.580 g*

* ! mol = x g*

* *

*x g = 0.580 g *

* 1.32 *x*10 ^{2}mol*

* *

Do the following problems:

1.
Calculate the volume of 36.0 g of steam at 115 ^{o}C and 110.0
kPa of pressure.

2.
If 28 g of N_{2}, nitrogen gas, occupy 22.4 L at STP calculate
the volume of 7 g of N_{2} at 0 ^{o}C and 202.6 kPa of pressure.

3.
If 6 g of a gas occupy 20 L at 40 ^{o}C and 303.9 kPa of
pressure, find its molar mass.

4.
A tire with a volume of 3.60 L contains 0.367 moles of air at a pressure
of 252 kPa. What is the temperature, in ^{o}C, of the air in the tire.

**
Stoichiometric Calculations**

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- stoichiometric calculations make use of balanced equations to calculate quantities of reactants and products

Example:

Given the following balanced equation

3Cu_{(s)}
+ 8HNO_{3(aq)} →
3Cu(NO_{3})_{2(aq)} + 2NO_{(g)} + 4H_{2}O_{(l)}

determine the volume of 2.50 mol/L nitric acid that would be required to react with 15.5 g of copper.

__Solutions__

* *

*Use the balanced equations
and proceed to write mole ratio, mass ratio and volume ratio.*

* *

3Cu_{(s)}
+ 8HNO_{3(aq)} →
3Cu(NO_{3})_{2(aq)} + 2NO_{(g)} + 4H_{2}O_{(l)}

*mole ratio:
3 8 3
2 4*

* *

*mass ratio: 190.65
g 504.16 g 562.71 g 60.02 g 72.08 g*

* *

*vol. ratio
44.8 L*

*(STP)*

* *

* convert 15.5 g
of copper to moles*

* moles of Cu =
15.5 g x 63.55 g/mol = 0.244 moles*

* *

* Determine
the number of moles of HNO_{3}*

* from the
equation we see that 3 moles Cu →
8 mole HNO _{3}*

*
0.244 moles Cu →x
moles HNO _{3}*

* *

*
x = 8 moles HNO_{3}
*

*
3 moles Cu*

* *

* Determine
the volume of acid, HNO_{3}*

* *

* Use the
formula c = n/v*

* *

* Find volume; v
= n/c*

*
= 0.651 moles/2.50mol/L*

*
= 0.260 L or 260 mL*

* *

Example:

Given the balanced equation

3Cu_{(s)}
+ 8HNO_{3(aq)} →
3Cu(NO_{3})_{2(aq)} + 2NO_{(g)} + 4H_{2}O_{(l)}

determine the mass of copper
that would be required to produce 4.00 L of nitrogen monoxide at 102.5 kPa and
22 ^{o}C.

__Solution:__

__
__

*Determine the number of
moles of NO _{(g)}*

*Rearrange the ideal gas
law equation*

* *

* n = PV*

*
RT*

* *

* n = _ 102.5
kPa *

* 8.314 kPa
L mol ^{−1}K^{−1}*x

* *

*From the balanced equation
we see that *

* *

*3 moles Cu → 2 moles NO
produced*

* *

*Determine the number of
moles of Cu*

* *

* Moles Cu =
0.167 moles NO *

*
2 mol NO*

*The mass of copper is*

* *

* Mass of Cu =
0.251 mol Cu *

*
1 mol Cu*

* *

- If measured amounts of two substances are reacted then we can calculate the number of moles of

each reagent used

- then the balanced equation can be used to determine which material is the limiting reagent

- the quantity of product will depend on the limiting reagent

__
__

Sample calculation:

Copper reacts with dilute nitric acid according to the equation:

3Cu_{(s)}
+ 8HNO_{3(aq)} →
3Cu(NO_{3})_{2(aq)} + 2NO_{(g)} + 4H_{2}O_{(l)}

A 200.0 mL sample of 2.00
mol/L nitric acid is added to 30.48 g of copper. Determine which reactant is the
limiting reagent. What volume of nitrogen monoxide, measured at 99.5 kPa and 20
^{o}C, would be produced?

* solution*

*determine the # of
moles of acid*

* *

*moles of HNO _{3} =
0.2000 L HNO_{3} *x

*
= 0.400 mol HNO _{3}*

* *

*determine the number of
moles of copper*

* *

*moles of Cu = 30.48 g*
χ * 63.55 g/mol = 0.4796 mol Cu*

* *

*determine the limiting
reagent *

* *

*from the equation we see
that*

*
3 mol Cu
→ 8 mol HNO _{3}
using # mole Cu actually have,*

*
0.4796 mol Cu
→ x mol HNO _{3}
calculate the # of moles of HNO_{3}
required*

* *

*
x mol HNO _{3} = 8 mol
HNO_{3} *

*3 mol Cu*

*
= 1.279 mol
HNO _{3}*

* *

*only 0.400 mol of HNO _{3}
are available, therefore the limiting reagent is HNO_{3}*

* *

*determine the # of
moles of NO*

* *

*the amount of limiting
reagent 0.400 mol HNO _{3} determines the amount of NO produced *

*from the balanced equation
we know that the ratio is*

* *

*
8 mol HNO _{3}
→ 2 mol NO produced*

*
0.400 mol HNO _{3}
→ x mol NO *

* *

*
x = 2 mol NO *

*
8 mol HNO _{3}*

*
= 0.100 mol NO*

* *

*determine the volume of
NO by using the gas law*

* *

*V = nRT*

* P*

*V = 0.100 mol *

*
99.5 kPa*

= *2.45 L NO*

* *

**
Stoichiometric** calculations determine the quantity of product that
will form in a reaction ; this represents the maximum possible yield or what is
referred to as the **theoretical yield**

when we carry out a reaction
in the laboratory we often obtain less than the theoretical yield; this is the
**actual yield**. The **percent yield** indicates how much product we
produced as a percentage. This is calculated by the expression:

percent yield = __actual yield__
x 100%

theoretical yield

Sample calculation:

Copper (II) oxide may be reduced to copper by heating it in a stream of hydrogen gas:

CuO_{(s)}
+ H_{2(g)} → Cu_{(s)}
+ H_{2}O_{(g)}

In an experiment 13.65 g of copper(II) oxide produces 10.75 g of copper metal. What is the percent yield?

__solution:__

* *

*determine the number of
moles of copper(II) oxide*

*moles of CuO = 13.65 g CuO
*χ *79.55 g/mol CuO*

* =
0.1716 mol CuO*

* *

*determine the number of
moles of Cu*

*from the equation we see
that 1 mol CuO → 1 mol Cu*

*
therefore 0.1716 mol CuO
→ 0.1716 mol Cu*

* *

*determine the possible
yield of Cu (theoretical yield)*

*mass of Cu = 0.1716 mol Cu
*x* 63.55 g/mol*

* = 10.91
g Cu*

* *

*determine the percent
yield*

*percent yield = 10.75 g
Cu *x

*
10.91 g Cu *

* *

1.
Red lead is an oxide of lead that has the formula Pb_{3}O_{4}.
When boiled with dilute nitric acid, the red lead reacts to produce lead(II)
nitrate and lead(IV) oxide:

Pb_{3}O_{4(s)} + 4HNO_{3(aq)}→
2Pb(NO_{3})_{2(aq) } + PbO_{2(s)} + 2H_{2}O(l)

a) How many moles of lead(II) nitrate would be produced from the complete reaction of 0.46 mol of red lead with an excess of nitric acid?

b) What volume of 2.50 mol/L nitric acid would be required to react with 15.85 g of red lead.

c) What mass of red lead would be required to react to producer 14.75 g of lead ((IV) oxide?

d) What is the percent yield if 11.25 g of red lead reacts with an excess of nitric acid to produce 3.85 g of lead(IV) oxide?

2. Bromic acid will oxidize sulfur dioxide to sulfuric acid

2HBrO_{3(aq)} + 5SO_{2(g)} + 4H_{2}O_{(l)}
→ Br_{2(aq)} + 5H_{2}SO_{4(aq)}

In an experiment 0.100 L of 0.200 mol/L bromic acid is reacted with 0.100L of a solution of sulfur dioxide.

a)
What volume of sulfur dioxide measured at 23 ^{o}C and 98.7 kPa
would neede to be dissolved in the aqueous solution to react will all the bromic
acid?

b) What is the maximum number of moles of sulfuric acid that could be produced?

c) Calculate the percent yield of sulfuric acid if the concentration of the solution produced was found to be 0.230 mol/L?

3. Aluminum metal can displace silver from a solution of silver nitrate. In an experiment 0.270 g of aluminum is added to 40.0 mL of 1.00 mol/L silver nitrate solution.

a) Write a balanced equation for the reaction.

b) Determine which of the reactants is the limiting reagent.

c) What is the theoretical yield of silver?

d) The mass of silver collected in the experiment was 2.98 g. Calculate the percent yield.

OTHER PROBLEMS:

1. How many atoms are there in one mole of oxygen gas?

2. What is the mass of

a) 238 moles of iron metal

b)
0.0035 moles of Na_{2}HPO_{4}

c)
7.70 x 10^{−6} moles of helium gas

d)
1.206 x 10^{24} atoms of N_{2} gas

3. How many moles are contained in the following substances

a) 1.50 g of silver metal

b) a 127 g spool of copper wire

c) 4.00 g sample of argon gas

4. If potassium chlorate is heated gently, the crystals will melt. Further heating will decompose it to give oxygen gas and potassium chloride.

a) Write the formula for the reactants and products.

b) Write the balanced equation for the decomposition of potassium chlorate.

c) How many grams of potassium chloride are produced when 3.5 moles of potassium chlorate are decomposed?

d) How many moles of potassium chlorate are needed to give 1.5 moles of oxygen gas.

e) How many liters of oxygen gas at STP will be produced by decomposing 122.6 g of potassium chlorate?

5.
Xenon is a noble gas. One of the first compounds made from xenon was
xenon tetrafluoride, XeF_{4}. When 10.0 g of xenon gas react completely
with fluorine to give xenon tetrafluoide, 39.8 kJ of energy are released. The
equation for the reaction is:

Xe_{(g)} + 2F_{2(g)} → XeF_{4(s)}

Using this information calculate how much energy is released when 1 mole of xenon gas reacts.

6a) How would you go about
preparing 150 mL of 1.5 M barium nitrate solution from solid barium nitrate,
Ba(NO_{3})_{2}?

b) Determine the molarity of a solution containing 0.15 g of KOH in 25 mL of solution.

6.
Solutions of FeBr_{3} and KOH are mixed and a reddish solid,
called a precipitate, Fe(OH)_{3}, is formed.

a) Write the molecular equation for the above reaction. It must be balanced.

b) Write the total ionic equation for the reaction.

c) Write the net ionic equation for the reaction.