basic stoichiometry

Basic Stoichiometry

Pisgah High SchoolM. Jones

Revision history: 5/16/03, 02/04/12, 04/27/12

With the Gas Laws

The word stoichiometry comes from the Greek words stoicheion which means “element” and metron which means “measure”.

Stoichiometry deals with the amounts of reactants and products in a chemical reaction.

Stoichiometry deals with moles.

Recall that …

1 mole = 22.4 L of any gas at STP

1 mole = the molar mass

molecules1 mole = 6.022 x 1023 atoms or

The word mole is one that represents a very large number.

… “mole” means 6.022 x 1023

Much like “dozen” means 12,

The key to doing stoichiometry is the balanced chemical equation.

2 H2 + O2 2 H2O22

The coefficients give the relative number of atoms or molecules of each reactant or product …as well as the number

of moles.

2 H2 + O2 2 H2O

2 H2 + O2 2 H2O2 moleculesof hydrogen

1 molecule of oxygen

2 moleculesof water

Two molecules of hydrogen combine with one molecule of oxygen to make two molecules of water.

2 H2 + O2 2 H2O

The balanced chemical equation also gives the smallest integer number of moles.

2 moleculesof hydrogen

1 molecule of oxygen

2 moleculesof water

2 H2 + O2 2 H2O

The balanced chemical equation also gives the smallest integer number of moles.

2 molesof hydrogen

1 moleof oxygen

2 molesof water

2 H2 + O2 2 H2O2 molesof hydrogen

1 moleof oxygen

2 molesof water

Two moles of hydrogen combine with

one mole of oxygen to make two moles of water.

Applications of Gay-Lussac’s Law

Simply put, Gay-Lussac’s Law says this:

The volumes of the gases are in the same ratio as the coefficients in the balanced equation.

2 H2 + O2 2 H2O2 moles 1 mole 2 moles2 L 1 L 2 L

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

1 mol 5 mols 3 mols 4 mols

Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required?

17.5 L O2

1 L 5 L 3 L 4 L

HC L 1

O L 5 HC L .53

83

283

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

1 mol 5 mols 3 mols 4 mols

Suppose 3.5 L of propane gas at STP is burned in oxygen, how many L at STP of oxygen will be required? 17.5 L O2

How many liters of carbon dioxide gas and water vapor at STP would be produced?

10.5 L CO2 and 14.0 L H2O

1 L 5 L 3 L 4 L

CH4(g) + 4 Cl2(g) CCl4(g) + 4 HCl(g)

When methane gas is allowed to react with an excess of chlorine gas, tetrachloromethane and hydrogen chloride gas will be produced.

How many L of methane will react with 0.800 L of chlorine gas at STP?

Cl L 4

CH L 1 Cl L 800.0

2

42 0.200 L Cl2

Stoichiometry problems involving gases

Find the mass of HOCl that can be produced when 2.80 L of chlorine gas at STP reacts with excess hydrogen peroxide.

Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g

Convert 2.80 L of Cl2 gas at STP to moles

Cl2(g) + H2O2 (l) 2 HOCl (l)2.80 L ??? g

Cl L 22.4

Cl mol 1 Cl L .802

2

22 0.125 mol Cl2

0.125 mol.125 x 2

0.250 mol

HOCl mol 1

HOCl g 52.5 HOCl mol 250.0 13.1g

HOCl

The reaction between copper and nitric acid

Will copper dissolve in acids?

Cu + 2HCl CuCl2 + 2H2 (g) No Reaction

Most metals react with HCl to produce a metal chloride solution and H2 gas.

Not copper

Consider hydrochloric acid

Will copper dissolve in acids?

Cu + 2HCl CuCl2 + 2H2 (g) No Reaction

Consider hydrochloric acid

Copper is below hydrogen in the activity series. Copper metal will only replace elements that are below it in the activity series.

What about other acids?

The same is true for all acids except nitric acid

Cu + HBr NRCu + HI NRCu + HF NRCu + H2SO4 NRCu + HC2H3O2 NR

Nitric acid is the only acid that will dissolve copper.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

A beaker contains a penny and some nitric acid is added.


The penny begins to disappear and the solution turns blue-green and a brown gas is given off.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O


The gas produced in the reaction is NO, which is colorless. Reddish brown NO2 forms when NO reacts with the oxygen in the air.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O


The penny is gone and the solution turns a dark blue. The brown NO2 gas escapes from the open beaker.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O


Calculate the volume of NO gas at STP when 20.0 grams of copper dissolves.

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O


20.0 g ??? L

Cu g 63.5

Cu mol 1Cu g 0.20 0.315 mol Cu

0.315 mol 0.210 mol x 0.667

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O


20.0 g

NO mol 1

NO L 22.4 NO mol 0.210 4.70 L NO

0.315 mol

??? L

0.210 mol

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

x 0.667

Using the Combined Gas Law

What if the conditions are not at STP?

What will be the volume of the NO gas at room temperature

in the mountains?

The temperature is 25 C, and the pressure is 691 torr.


There are two possible solutions.

1. Use the Combined Gas Law to compute the new volume at the new temperature and pressure.

2

22

1

11

T

VP

T

VP


20.0 g

NO mol 1

NO L 22.4 NO mol 0.210 4.70 L NO

0.315 mol

??? L

0.210 mol

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

x 0.667

What is the volume at a temperature of 25 C, and a pressure of 691 torr?

20.0 g 4.70 L (STP)

Using the Combined Gas Law

2

22

1

11

T

VP

T

VP

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

2

22

1

11

T

VP

T

VPSolve for V2

21

2112 PT

TVPV

torr691K 273

K 298L 70.4torr607

V2 = 5.65 L

V2 is the new volume at 25 C and 691 torr

Using the Ideal Gas Equation


There are two possible solutions.

2. Use the Ideal Gas Equation to compute the new volume using the new temperature and pressure.

nRTPV

nRTPV Moles of gas

Pressure

Volume in Liters

The gas constant

Temperature in K

The value of R depends on pressure units.

20.0 g

NO mol 1

NO L 22.4 NO mol 0.210 4.70 L NO

0.315 mol

??? L

0.210 mol

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

x 0.667

We know that 0.210 mol of NO gas at STP was produced from 20.0g Cu.

What is the volume at a temperature of 25 C, and a pressure of 691 torr?

20.0 g 0.210 mol

Using the Ideal Gas Eauation

PV = nRT

3Cu + 8HNO3 3Cu(NO3)2 + 2NO + 4H2O

Solve PV=nRT for V

P

nRT V

torr 691

K 298 62.4 mol 0.210 V K mol

torrL

Then plug in the new temperature and pressure.

V = 5.65 L

Either method works fine in this case, since you have a gas that you know both the pressure, and the number of moles.

Use the Combined Gas Law when you are looking for a new pressure, volume or temperature for a confined gas. (moles are constant)

Use the Ideal Gas Equation when you are looking for a pressure, volume or temperature when the number of moles is known.

Of course the Ideal Gas Equation can be used whenever you are dealing with pressure, volume, temperature or the number of moles, and any one of the variables could be the unknown.

Practice solving PV=nRT for each variable.

Remember, when using PV = nRT …

Pressure is usually in atmospheres, torr, mm Hg, or kilopascals.

Volume is usually in liters.… and n is in moles.

The value of R, the gas constant, depends on the units of V and P.

Temperature must be in Kelvin…

Simple ideal gas problem:

How many moles are in a 2.00 L container of oxygen gas at a pressure of 0.950 atm and a temperature of 20.0 C?

This could actually be part of a larger problem:

Find the mass of iron(III) oxide formed when excess iron reacts with 2.00 L of oxygen gas at 0.950 atm and a temperature of 20.0 C.

Start with the balanced equation:

4Fe(s) + 3O2(g) 2Fe2O3(s)Excess 0.950 atm, 2.00 L, 20.C

Use the ideal gas law to find the moles of O2.

PV = nRT

K2930821.0

L00.2atm950.0

RT

PVn

molK

Latm

= 0.0790 mol

Start with the balanced equation:

4Fe(s) + 3O2(g) 2Fe2O3(s)Excess 0.950 atm,

2.00 L, 20.C 0.0790 mol 0.0527 mol

8.41 g

x 2/3

32

3232 OFemol1

OFeg6.159OFemol0527.0 = 8.41g Fe2O3

There is a useful variation on the Ideal Gas Equation when dealing with molar mass. Since …

Molar mass

mass n

M

m

Then PV=nRT becomes M

mRT PV

An example of using

M

mRT PV

Find the volume of 0.325 grams of nitrogen gas at a pressure of 1.20 atm and a temperature of 300. K.

M

mRT PV

Use this equation because you are given the mass of a gas for which you can easily calculate the molar mass.

Find the volume of 0.325 grams of nitrogen gas at a pressure of 1.20 atm and a temperature of 300. K.

P

mRT V

M

atm 1.20 0.28

300.K0821.0 g 0.325

mol

gK mol

atm L

V = 0.238 L

Solve for V

Look forward to more stoichiometry problems using

the Ideal Gas Equation.

basic stoichiometry

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