The
quantitative relationship among reactants and products is called stoichiometry.
The term stoichiometry is derived from two Greek words: stoicheion (meaning
"element") and metron (meaning "measure"). On this subject,
you often are required to calculate quantities of reactants or products.
Stoichiometry
calculations are based on the fact that atoms are conserved. They cannot be
destroyed or created. Numbers and kinds of atoms before and after the reactions
are always the same. This is the basic law of nature. From the atomic and molecular point of view,
the stoichiometry in a chemical reaction is very simple. However, atoms of
different elements and molecules of different substances have different
weights. Thus, simple stoichiometry at the atomic level appears to be
complicated when amounts (quantities) are measured in units of g, kg, L or mL.
When quantities in moles are used, the relationships (or ratios) are really
simple. For example, one mole of oxygen reacts with two moles of hydrogen,
2 H2
+ O2 -> 2 H2O
or one mole of hydrogen reacts with
half a mole of oxygen,
H2
+ ½ O2 -> H2O
one mole of carbon reacts with one
mole of oxygen.
C + O2
-> CO2
This
is a major and important topic that you have to master. In order to accomplish
this, you have to be able to do several things. First, you have to be able to
convert amounts of substances between mass units of g (or kg) to moles and vice
versa. Then, you have to understand chemical reactions (changes). In this case, you not only know what are the
reactants and products, you can write a balanced equation to explain the
reaction. Sometimes you may be told what the reactions are. There are many
chemical reactions, but they can be divided into a few types as a summary.
In
a chemical reaction, not all reactants are necessarily consumed. One of the
reactants may be in excess and the other may be limited. The reactant that is
completely consumed is called limiting reactant, whereas unreacted reactants are called excess reactants.
Amounts
of substances produced are called yields. The amounts calculated according to stoichiometry are
called theoretical yields whereas the actual amounts are called actual
yields. The actual yields are often expressed in percentage, and they are
often called percent yields. Listed
with the skills are some general questions. If you are confident in answering
all these questions, you have already acquired the skills. Furthermore, we list
some diagnostic problems below. Each question test some aspect of
stoichiometry. If you find a question difficult, the appropriate topic to study
is suggested.
Balancing
In
, chemical reactions are frequently written as an equation, using chemical
symbols. The reactants are displayed on the left side of the equation and the
products are shown on the right, with the separation of either a single or
double arrow that signifies the direction of the reaction. The significance of
single and double arrow is important when discussing solubility constants, but
we will not go into detail about it in this module. To balance an equation, it
is necessary that there are the same number of atoms on the left side of the
equation as the right. One can do this by raising the coefficients.
Reactants to Products
A
chemical equation is like a recipe for a reaction so it displays all the
ingredients or terms of a chemical reaction. It includes the elements,
molecules, or ions in the reactants and in the products as well as their
states, and the proportion for how much of each particle is create relative to
one another, through the stoichiometric coefficient. The following equation
demonstrates the typical format of a chemical equation:
2Na(s) + 2HCl(aq)
→ 2NaCl(aq) + H2(g)
In
the above equation, the elements present in the reaction are represented by
their chemical symbols. Based on the Law of Conservation
of Mass,
which states that matter is neither created nor destroyed in a chemical
reaction, every chemical reaction has the same elements in its reactants and
products, though the elements they are paired up with often change in a
reaction. In this reaction, sodium (Na), hydrogen (H), and
chloride (Cl) are the elements present in both reactants, so based on
the law of conservation of mass, they are also present on the product side of
the equations. Displaying each element is important when using the chemical
equation to convert between elements.
Stoichiometric Coefficients
In
a balanced reaction, both sides of the equation have the same number of
elements. The stoichiometric coefficient is the number written in front of
atoms, ion and molecules in a chemical reaction to balance the number of each
element on both the reactant and product sides of the equation. Though the
stoichiometric coefficients can be fractions, whole numbers are frequently used
and often preferred. This stoichiometric coefficients are useful since they
establish the mole ratio between reactants and products. In the balanced
equation:
2Na(s) + 2HCl(aq)
→ 2NaCl(aq)+H2(g)
we
can determine that 2 moles of HCl will react with 2 moles of Na(s)
to form 2 moles of NaCl(aq) and 1 mole of H2(g).
If we know how many moles of Na we start out with, we can use the ratio
of 2 moles of NaCl to 2 moles of Na to determine how many moles of NaCl
were produced or we can use the ration of 1 mole of H2 to 2 moles of Na
to convert to NaCl. This is known as the coefficient factor. The
balanced equation makes it possible to convert information about one reactant
or product to quantitative data about another element. Understanding this is
essential to solving stoichiometric problems.
Example 1
Lead
(IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.
Pb(OH)4 + H2SO4
→ Pb(SO4)2 + H2O
SOLUTION
Start
by counting the number of atoms of each element.
UNBALANCED
Element
|
Reactant
(# of atoms)
|
Product
(# of atoms)
|
Pb
|
1
|
1
|
O
|
8
|
9
|
H
|
6
|
2
|
S
|
1
|
2
|
The
reaction is not balanced; the reaction has 16 reactant atoms and only 14 product
atoms and does not obey the conservation of mass principle. Stoichiometric
coefficients must be added to make the equation balanced. In this example,
there are only one sulfur atom present on the reactant side, so a coefficient
of 2 should be added in front of H2SO4 to have an equal number of sulfur on both sides of the
equation. Since there are 12 oxygen on the reactant side and only 9 on the
product side, a 4 coefficient should be added in front of H2O where there is a deficiency of oxygen. Count the number of
elements now present on either side of the equation. Since the numbers are the
same, the equation is now balanced.
Pb(OH)4
+ 2H2SO4
→ Pb(SO4)2+4H2O
BALANCED
Element
|
Reactant
(# of atoms)
|
Product
(# of atoms)
|
Pb
|
1
|
1
|
O
|
12
|
12
|
H
|
8
|
8
|
S
|
2
|
2
|
Balancing
reactions involves finding least common multiples between numbers of elements
present on both sides of the equation. In general, when applying coefficients,
add coefficients to the molecules or unpaired elements last.
A balanced equation ultimately has
to satisfy two conditions.
- The numbers of each element on the left and right side of the equation must be equal.
- The charge on both sides of the equation must be equal. It is especially important to pay attention to charge when balancing redox reactions.
Stoichiometry
and Balanced Equations
In
stoichiometry, balanced equations make it possible to compare different
elements through the stoichiometric factor discussed earlier. This is
the mole ratio between two factors in a chemical reaction found through the
ratio of stoichiometric coefficients. Here is a real world example to show how
stoichiometric factors are useful.
Example 2
There
are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to
be sent. How many party invitations can be sent?
SOLUTION
The
equation for this can be written as
I +
2S → IS2
where
:
- I represents invitations,
- S represents stamps, and
- IS2 represents the sent party invitations consisting of one invitation and two stamps.
Based
on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced
equation. Invitations Stamps Party Invitations Sent
In
this example are all the reactants (stamps and invitations) used up? No, and
this is normally the case with chemical reactions. There is often excess of one
of the reactants. The limiting reagent, the one that runs out first, prevents
the reaction from continuing and determines the maximum amount of product that
can be formed.
Example 3
SOLUTION
Stamps, because there was only
enough to send out invitations, whereas there were enough invitations for 12
complete party invitations. Aside from just looking at the problem, the problem
can be solved using stoichiometric factors.
12
I x (1IS2/1I) = 12 IS2 possible
20
S x (1IS2/2S) = 10 IS2 possible
When there is no limiting reagent
because the ratio of all the reactants caused them to run out at the same time,
it is known as stoichiometric proportions.
Types
of Reactions
There are 6 basic types of
reactions.
- Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical and O2
- Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a complex product.
- Decomposition: Decomposition is when complex reactants are broken down into simpler products.
- Single Displacement: Single displacement is when an element from on reactant switches with an element of the other to form two new reactants.
- Double Displacement: Double displacement is when two elements from on reactants switched with two elements of the other to form two new reactants.
- Acid-Base: Acid- base reactions are when two reactants form salts and water.
Molar
Mass
Before
applying stoichiometric factors to chemical equations, you need to understand
molar mass. Molar mass is a useful chemical ratio between mass and moles. The
atomic mass of each individual element as listed in the periodic table
established this relationship for atoms or ions. For compounds or molecules,
you have to take the sum of the atomic mass times the number of each atom
in order to determine the molar mass
Example 4
What is the molar mass of H2O?
SOLUTION
Molar mass=2×(1.00794g/mol)+1×(15.9994g/mol)=18.01528g/mol
Using molar mass and coefficient
factors, it is possible to convert mass of reactants to mass of products or
vice versa.
Example 5:
Combustion of Propane
Propane (C3H8) burns in this reaction:
C3H8+5O2→4H2O+3CO2
If 200 g of propane is burned, how
many g of H2O is produced?
SOLUTION
SOLUTION
Steps
to getting this answer: Since you cannot calculate from grams of reactant to
grams of products you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O. Then convert from moles of H2O to grams of H2O.
- Step 1: 200 g C3H8 is equal to 4.54 mol C3H8 .
- Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol H2O.
- Step 3: Convert 18.18 mol H2O to g H2O. 18.18 mol H2O is equal to 327.27 g H2O.
Variation
in Stoichiometric Equations
Almost every quantitative
relationship can be converted into a ratio that can be useful in data analysis.
Density
Density
(ρ) is calculated as mass/volume. This
ratio can be useful in determining the volume of a solution, given the mass or
useful in finding the mass given the volume. In the latter case, the inverse
relationship would be used.
Volume
x (Mass/Volume) = Mass
Mass
x (Volume/Mass) = Volume
Percent
Mass
Percents
establish a relationship as well. A percent mass states how many grams of a
mixture are of a certain element or molecule. The percent X% states that of
every 100 grams of a mixture, X grams are of the stated element or compound.
This is useful in determining mass of a desired substance in a molecule.
Example 6
A
substance is 5% carbon by mass. If the total mass of the substance is 10 grams,
what is the mass of carbon in the sample? How many moles of carbon are there?
SOLUTION
SOLUTION
10
g sample x (5 g carbon/100 g sample) = 0.5 g carbon
0.5g
carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon
Molarity
Molarity
(moles/L) establishes a relationship between moles and liters. Given volume and
molarity, it is possible to calculate mole or use moles and molarity to
calculate volume. This is useful in chemical equations and dilutions.
Example 7
How much 5 M stock solution is
needed to prepare 100 mL of 2 M solution?
SOLUTION
100 mL of dilute solution (1 L/1000
mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock
solution/1L stock solution) = 40 mL stock solution.
These ratios of molarity, density,
and mass percent are useful in complex examples ahead.
Determining
Empirical Formulas
An
empirical formula can be determined through chemical stoichiometry by
determining which elements are present in the molecule and in what ratio. The
ratio of elements is determined by comparing the number of moles of each
element present.
Example 8:
Combustion of Organic Molecules
1.000
gram of an organic molecule burns completely in the presence of excess oxygen.
It yields 0.0333 mol of CO2 and 0.599 g of H2O. What is
the empirical formula of the organic molecule?
SOLUTION
This
is a combustion reaction. The problem requires that you know that organic
molecules consist of some combination of carbon, hydrogen, and oxygen elements.
With that in mind, write the chemical equation out, replacing unknown numbers
with variables. Do not worry about coefficients here.
CxHyOz(g)+O2(g)→CO2(g)+H2O(g)
Since
all the moles of C and H in CO2 and H2O, respectively
have to have came from the 1 gram sample of unknown, start by calculating how
many moles of each element were present in the unknown sample.
0.0333mol
CO2 (1mol C/ 1mol CO2) = 0.0333mol C in unknown
0.599g
H2O (1mol H2O/ 18.01528g H2O)(2mol H/ 1mol H2O)
= 0.0665 mol H in unknown
Calculate
the final moles of oxygen by taking the sum of the moles of oxygen in CO2
and H2O. This will give you the number of moles from both the
unknown organic molecule and the O2 so you must subtract the moles
of oxygen transferred from the O2.
Moles of oxygen in CO2:
0.0333mol
CO2 (2mol O/1mol CO2) = 0.0666 mol O
Moles of oxygen in H2O:
0.599g
H2O (1mol H2O/18.01528 g H2O)(1mol O/1mol H2O)
= 0.0332 mol O
Using
the Law of Conservation, we know that the mass before a reaction must equal the
mass after a reaction. With this we can use the difference of the final mass of
products and initial mass of the unknown organic molecule to determine the mass
of the O2 reactant.
0.333mol
CO2(44.0098g CO2/ 1mol CO2) = 1.466g CO2
1.466g
CO2 + 0.599g H2O - 1.000g unknown organic = 1.065g O2
Moles of oxygen in O2
1.065g
O2(1mol O2/ 31.9988g O2)(2mol O/1mol O2)
= 0.0666mol O
Moles of oxygen in unknown
(0.0666mol
O + 0.0332 mol O) - 0.0666mol O = 0.0332 mol O
Construct a mole ratio for C, H, and
O in the unknown and divide by the smallest number.
(1/0.0332)(0.0333mol
C : 0.0665mol H : 0.0332 mol O) => 1mol C: 2 mol H: 1 mol O
From this ratio, the empirical formula
is calculated to be CH2O.
Determining
Molecular Formulas
To
determine a molecular formula, first determine the empirical formula for the
compound as shown in the section above and then determine the molecular mass
experimentally. Next, divide the molecular mass by the molar mass of the
empirical formula (calculated by finding the sum the total atomic masses of all
the elements in the empirical formula). Multiply the subscripts of the
molecular formula by this answer to get the molecular formula.
Example 9
In the example above, it was
determined that the unknown molecule had an empirical formula of CH2O.
1. Find the molar mass of the
empircal formula CH2O.
12.011g
C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH2O
2. Determine the molecular mass
experimentally. For our compound, it is 120.056 g/mol.
3. Divide the experimentally
determined molecular mass by the mass of the empirical formula.
(120.056
g/mol) / (30.026 g/mol) = 3.9984
4.
Since 3.9984 is very close to four, it is possible to safely round up and
assume that there was a slight error in the experimentally determined molecular
mass. If the answer is not close to a whole number, there was either an error
in the calculation of the empirical formula or a large error in the
determination of the molecular mass.
5. Multiply the ratio from step 4 by
the subscripts of the empirical formula to get the molecular formula.
CH2O * 4 = ?
C: 1 * 4 = 4
H: 2 * 4 = 8
O 1 * 4 = 4
CH2O * 4 = C4H8O4
6. Check your result by calculating
the molar mass of the molecular formula and comparing it to the experimentally
determined mass.
molar mass of C4H8O4=
120.104 g/mol
experimentally determined mass =
120.056 g/mol
% error = | theoretical -
experimental | / theoretical * 100%
% error = | 120.104 g/mol -
120.056 g/mol | / 120.104 g/mol * 100%
% error = 0.040 %
Example 10:
Complex Stoichiometry Problem
An
amateur welder melts down two metals to make an alloy that is 45% copper by
mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of
alloy completely fills a mold of volume 1000 cm3. He accidently
breaks off a 1.203 cm3 piece of the homogenous mixture and
sweeps it outside where it reacts with acid rain over years. Assuming the acid
reacts with all the iron(II) and not with teh copper, how many grams of H2(g)
are released into the atmosphere because of the amateur's carelessness? (Note
that the situation is fiction.)
SOLUTION
Step 1: Write a balanced equation after determining the products
and reactants. In this situation, since we assume copper does not react, the
reactants are only H+(aq) and Fe(s). The given product is H2(g) and
based on knowledge of redox reactions, the other product must be Fe2+(aq).
Fe(s)+2H+(aq)→H2(g)+Fe2+(aq)
Step 2: Write down all the given information
Alloy
density = (3.15g alloy/ 1L alloy)
x
grams of alloy = 45% copper = (45g Cu(s)/100g alloy)
x
grams of alloy = 55% iron(II) = (55g Fe(s)/100g alloy)
1 liter alloy = 1000cm3 alloy
alloy sample = 1.203cm3 alloy
Step 3: Answer the question of what is being asked. The
question asks how much H2(g) was produced. You are expected to solve for the
amount of product formed.
Step 4: Start with the compound you know the most about and
use given ratios to convert it to the desired compound.
Convert
the given amount of alloy reactant to solve for the moles of Fe(s) reacted.
1.203cm3 alloy(1liter
alloy/1000cm3 alloy)(3.15g alloy/1liter alloy)(55g Fe(s)/100g
alloy)(1mol Fe(s)/55.8g Fe(s))=3.74 x 10-5 mol Fe(s)
Make sure all the units cancel out
to give you moles of Fe(s). The above conversion involves using
multiple stoichiometric relationships from density, percent mass, and
molar mass.
The balanced equation must now be
used to convert moles of Fe(s) to moles of H2(g). Remember that the
balanced equation's coeffiecients state the stoichiometric factor or
mole ratio of reactants and products.
3.74 x 10-5 mol Fe
(s) (1mol H2(g)/1mol Fe(s)) = 3.74 x 10-5 mol H2(g)
Step 5: Check units
The question asks for how many grams of
H2(g) were released so the moles of H2(g) must still be
converted to grams using the molar mass of H2(g). Since there are
two H in each H2, its molar mass is twice that of a single H atom.
molar
mass = 2(1.00794g/mol) = 2.01588g/mol
3.74 x 10-5 mol H2(g)
(2.01588g H2(g)/1mol H2 (g)) = 7.53 x 10-5 g
H2(g) released
Problems
Stoichiometry
and balanced equations make it possible to use one piece of information to
calculate another. There are countless ways stoichiometry can be used in and
everyday life. Try and see if you can use what you learned to solve the
following problems.
1) Why are the following equations
not considered balanced?
- H2O(l)→H2(g)+O2(g)
- Zn(s)+Au+(aq)→Zn2+(aq)+Ag(s)
2) Hydrochloric acid reacts with a
solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the
balanced chemical equation for this reaction.
3) Given a 10.1M stock solution, how
many mL must be added to water to produce 200 mL of 5M solution?
4)
If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide
and water, what is the limiting reagent and how many moles of water are
produced? The unbalanced equation is provided below.
CH4(g)+O2(g)→CO2(g)+H2O(l)
5)
A 0.777g sample of an organic compound is burned completely. It produces 1.42g
CO2 and 0.388g H2O. Knowing that all the carbon and
hydrogen atoms in CO2 and H2O came from the 0.777g
sample, what is the empirical formula of the organic compound?
Source
:
http://www.science.uwaterloo.ca/~cchieh/cact/c120/stoichio.html
https://chem.libretexts.org/Core/Inorganic_Chemistry/Chemical_Reactions/Stoichiometry_and_Balancing_Reactions
What is a limiting reaction? Give a sample form of workmanship
BalasHapusWhen there is not enough of one reactant in a chemical reaction, the reaction stops abruptly. To figure out the amount of product produced, it must be determined reactant will limit the chemical reaction (the limiting reagent) and which reactant is in excess (the excess reagent). One way of finding the limiting reagent is by calculating the amount of product that can be formed by each reactant; the one that produces less product is the limiting reagent.
HapusAre u know about the laws that govern the stoichiometry? Please give explaination,thx
BalasHapusStoichiometry is the study of quantitative relationship among substances involved in chemical changes. This relationship, called the stoichiometry of a reaction, can be used to predict how much reactant is needed to create a certain amount of product or to predict how much of the product will be formed from a certain amount of reactant. It also defines the proportions in which reactants will combine and the proportion in which products will be formed. Underlying the study of stoichiometry are a number of principles, or , that provide the basis for the relationships among reactants and products in a chemical reaction.
HapusThe law of conservation of mass was first expounded by Russian chemist Mikhail Lomonosov (1711-1765) who investigated the reactions of compounds with air in closed containers. The mass--before and after the reaction--remained constant despite the observable change. Mass was neither being created nor destroyed in the reaction, but merely converted from one form to another. Unfortunately, working in Russia in the 1750s, Lomonosov's results did not get the attention or recognition that he most likely deserved. French chemist Antoine Laurent Lavoisier (1743-1794) is most often credited with the formulation of this law. Prior to the development of atomic theory, this law was, simply, the absolute statement that there is no measurable change in the total mass of chemical substances involved in a reaction.
The law of conservation of mass can be seen using the simple decomposition of water according to the reaction:
2H2O → 2H2 + O2.
The data for this experiment demonstrate that, for instance, 11.2 grams of hydrogen and 88.8 grams of oxygen are produced by 100 grams of water, regardless of the water's source. Furthermore, these elements can be converted back into 100 grams of water if the reaction is reverse by combining the two elements. Similarly, the burning of a candle, the rusting of a metal, or the act of living all obey the conservation law and this can be demonstrated, provided the reactions are carried out in a closed system to prevent the loss of gaseous products.
With the advent of the atomic theory, scientists soon realized that it is not just matter or mass that is conserved but atoms as well. The more modern formulation of the law of conservation of mass states that atoms can be neither created nor destroyed during a chemical reaction--only converted or rearranged into new substances. However, this law was pre-dated by the law of constant composition and the law of definite proportions. The first law of these laws states that the elemental composition of a compound is the same, regardless of the source of the compound. That is, water is always H2O, whether it is obtained from a snow pack of a Himalayan mountaintop or the bathroom tap. The second law is an alternative way of looking at the same concept: When elements combine to form a chemical compound, they do so in a definite proportion by mass. That is, all molecules of calcium carbonate are the same, consisting of a calcium ion and carbonate ion, and the carbonate ion is further composed of a carbon atom and three oxygen atoms. The composition is invariant; the mass proportions are fixed.
Please explain to me the basic principles of stoichiometric experiments?
BalasHapusokay thanks rahmi, the basic prinsiples of stoichiometri is...
HapusWe provided an overview of the most common mathematical techniques used in the stoichiometric analysis of metabolic networks. We have not described in any detail the application of these techniques to biological problems, which is found elsewhere [31–33]. These applications to biology are the reasons for the existence of pathway analysis, and there are a number of success stories [34, 35]. Yet, the simplifications and subsequent limitations of the described techniques are also clear and extensions to pathway analysis methods include the incorporation of dynamics (such as in dynamic FBA [36]), additional constraints (such as space or resource limitations [37–39], multidimensional optimality [30], and extensions to multi-species FBA [40–42]. It is therefore to be expected that such analysis will penetrate biology in increasingly many ways to provide rigorous and quantitative hypotheses and fundamental understanding.
"Numbers and kinds of atoms before and after the reactions are always the same" can you explain to me and give an example
BalasHapusokayrifaa,, thanks to the question the law of the stoichiometri are
Hapus1.Conservation of the Mass (Lavoisier)
"The mass of the reaction substance is equal to the mass of the substance after the reaction"
Example:
S (s) + O2 (g) → SO2 (g)
1 mol of S with 1 mol of O2 forming 1 mole of SO2. 32 grams of S with 32 grams of O2 formed 64 grams SO2. The total mass of the reactants is equal to the mass of the resulting product.
H2 (g) + ½ O2 (g) → H2O (l)
1 mole of H2 with ½ mole O2 forming 1 mole of H2O. 2 grams of H2 with 16 grams of O2 form 18 grams of H2O. The total mass of the reactants is equal to the mass of the product formed.
2. Equipment Fixed (Proust)
"The mass ratio of the constituent elements is always, whatever is made in different ways"
Example:
S (s) + O2 (g) → SO2 (g)
The way the mass of S to the mass of O2 to form SO2 is 32 grams of S to 32 grams of O2 or 1: 1. This means that every gram of S is exactly one gram of O2 forming 2 grams of SO2. If 50 grams of S is required, it takes 50 grams of O2 to form 100 grams of SO2.
H2 (g) + ½ O2 (g) → H2O (l)
One hundred mass of H2 to mass of O2 to form H2O is 2 gram H2 to 16 gram of O2 or 1: 8. This means, every one gram of H2 exactly with 8 gram O2 form 9 gram H2O. If given 24 grams of O2, it takes 3 grams of H2 to form 27 grams of H2O
3. Volume Comparison (Gay Lussac)
Applies only to chemical reactions that involve the gas phase
"At the same temperature and pressure, the ratio of reactant gas volume to the gas volume of the reaction product is a simple integer (equal to the ratio of the reaction coefficient)"
Example:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of N2 gas exactly reacts with 3 mL of H2 gas to form 2 mL of NH3 gas. Thus, to obtain 50 L of NH 3 gas, it takes 25 L of N2 gas and 75 L of H2 gas.
CO (g) + H2O (g) → CO2 (g) + H2 (g)
The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 mL of CO gas reacts exactly with 1 mL of H2O gas to form 1 mL of CO2 gas and 1 mL of H2 gas. Thus, as much as 4 L of CO gas requires 4 L of H2O gas to form 4 L of CO2 gas and 4 L of H2 gas.
4. Avogadro
Applies only to chemical reactions that involve the gas phase
"At the same temperature and pressure, the same volumes of gases contain the same number of moles"
Avogadro is closely related to Gay Lussac
Example:
N2 (g) + 3 H2 (g) → 2 NH3 (g)
The mole ratio is equal to the ratio of the reaction coefficient. This means that every 1 mole of precise N2 gas reacts with 3 moles of H 2 gas to form 2 moles of NH 3 gas. The gas volume ratio is equal to the ratio of the reaction coefficient. This means that every 1 L of N2 gas precisely reacts with 3 L of H 2 gas to form 2 L of NH3 gas. Thus, if at a certain temperature and pressure, 1 mole of gas is equivalent to 1 L of gas, then 2 moles of gas is equivalent to 2 L of gas. In other words, the mole gas ratio is equal to the ratio of gas volume.
Could you give me the best key for Solve stoichiometric problem?
BalasHapusStoichiometry is a section of that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balanced reactions.
HapusCould you give me some example for stoichiometry question and then the solution?
BalasHapusSTP is a temperature of 273 K and a pressure of 101,3 kPa. The amount of gas is usually 1 mol.
HapusWe write down all the values that we know about one mole of gas at STP:
Now we can substitute these values into the ideal gas equation:
pV(101\ 300)V(101\ 300)VVV=nRT=(1)(8,31)(273)=2265,9=0,0224 m3=22,4 dm3
The volume of 1 mole of gas at STP is 22,4 dm3.
And if we had any number of moles of gas, not just one mole then we would get:
Vg=22,4ng