Senin, 08 Mei 2017

Stoichiometry


Stoichiometry 


A. Understanding Stoichiometry
Stoichiometry comes from two Greek syllables Stoicheion meaning "element" and Metron which means "measurement".
Stoichiometry is a subject in chemistry involving the linkage of reactants and products in a chemical reaction to determine the quantity of each reacting agent.
Stoichiometry is a subject in chemistry that studies the quantity of matter in a chemical reaction.
In chemistry, stoichiometry (sometimes called stoichiometry of reaction to distinguish it from composition stoichiometry) is a science that studies and quantifies quantitative relationships of reactants and products in chemical reactions (chemical equations). This word comes from the Greek stoikheion (element) and metriā (size).

B. Chemical Reaction Equation

Chemical reactions are often written in bentu equations using element symbols. The reactants are the substances that are on the left, and the product is the substance that is on the right, then both are separated by arrows (can be one or two alternating arrows). Example:
2Na (s) + HCl (aq) → 2NaCl (aq) + H2 (g)
The equation of a chemical reaction is like a prescription in the reaction, thus indicating everything associated with the reaction, whether it is an ion, an element, a compound, a reactant or a product. All. Then as in the recipe, there is a proportion of the equation shown in the figures in front of the molecular formula.
When considered again, the number of H atoms on the reactant (left) is not the same as the number of H atoms on the product (right). Then this reaction needs to be synchronized. The equalization of chemical reactions must satisfy some chemical laws of matter.

C. The Law of Conservation of Mass

LAW MASS LAW = LAVOISIER LAW
"The mass of substances before and after the reaction is fixed".
Example:
Hydrogen + oxygen of hydrogen oxide
   
(4g) (32g) (36g)

D. Comparable Law (Proust Law)
Fixed Comparison Law: The chemical compound consists of chemical elements with the ratio of the mass of elements that remain the same.
PERSONAL COMPARATIVE LAW = PROUST LEGAL
"The ratio of the mass of the elements in each compound is fixed"

Example:
A. In the compound NH3: mass N: mass H
= 1 Ar. N: 3 Ar. H
= 1 (14): 3 (1) = 14: 3
B. In the SO3 compound: mass S: mass 0
= 1 Ar. S: 3 Ar. O
= 1 (32): 3 (16) = 32: 48 = 2: 3
Advantages of Proust Law:
If known mass of a compound or mass of one element that make up the compound make the mass of other elements can be known.
Example:
What is the level of C in 50 grams CaCO3? (Ar: C = 12, 0 = 16; Ca = 40)
Mass C = (Ar C / Mr CaCO3) x CaCO3 mass
= 12/100 x 50 grams = 6 grams
Mass C
Levels C = mass C / CaCO3 mass x 100%
= 6/50 x 100% = 12%

E. The Law of Multiple Comparisons (Dalton's Law)
The Law of Multiple Comparisons: If an element reacts with other elements, then the ratio of the weight of the element is a simple integer
 
So from persmaaan:
2Na (s) + 2HCl (aq) → 2NaCl (aq) + H2 (g)

We can know that 2 moles of HCl react with 2 moles of Na to form 2 moles of NaCl and 1 mole of H2. By equalizing this reaction, it can be known the quantity of each substance involved in the reaction.

Hence the equalization of this reaction is very important in solving stoichiometric problems.
Example:
Lead (IV) Hydroxide reacts with Sulfuric Acid, by reaction as follows:
Pb (OH)4 + H 2SO 4 → Pb (SO 4) 2 + H 2O.

 LEGAL COMPARATIVE LAW = LEGAL DALTON
"If two elements can form two or more compounds for the mass of one element equal to the number then the ratio of the mass of the second element will be proportional to the integer and the simple".
Example:
When the Nitrogen element of the oxygen dispensed can be formed,
NO where the mass N: 0 = 14: 16 = 7: 8
NO2 where the mass N: 0 = 14: 32 = 7: 16
For the same mass of mass Nitrogen the Oxygen mass ratio of NO: NO2 = 8: 16 = 1: 2

F. LAW-LAW GAS
For ideal gas apply equation: PV = nRT
Where:
P = gas pressure (atmosphere)
V = volume of gas (liter)
N = mol of gas
R = universal gas constant = 0.082 lt.atm / mol Kelvin
T = absolute temperature (Kelvin)
The changes of P, V and T from state 1 to state 2 under certain conditions are reflected by the following laws:

1.  BOYLE LAW
This law is derived from the ideal gas state equations with
N1 = n2 and T1 = T2; So obtained: P1.V1 = P2.V2

2. GAY-LUSSAC LAW
"The volume of the reacting gases and the volume of the gases of the bile reaction measured at the same temperature and pressure will be proportional to simple den".
So for: P1 = P2 and T1 = T2 apply: V1 / V2 = n1 / n2

3. BOYLE-GAY LUSSAC LAW
This law is an extension of the previous law den diturukan with the state of price n = n2 so that obtained the equation:
P1. V1 / T1 = P2. V2 / T2

4. . AVOGADRO LAW
"At the same temperature and pressure, the same volumes of gases contain the same number of moles. From this statement it is determined that in the STP state (0 ° C 1 atm) 1 mole per volume of 22.4 liters volumes this volume is referred to as the molar volume of the gas.
G. MASS ATOM AND MASS FORUM
1. Relative Atomic Mass (Ar)
Is the ratio between the mass of 1 atom with 1/12 mass 1 carbon atom 12
2. Relative Molecular Mass (Mr)
Is a comparison between the mass of 1 molecule of the compound with 1/12 mass 1 carbon atom 12.
The relative molecular mass (Mr) of a compound is the sum of the atomic mass of its constituent elements.
Example:
If Ar for X = 10 and Y = 50 what is Mr. compound X2Y4?
Answer:
Mr. X 2Y4 = 2 x Ar. X + 4 x Ar. Y = (2 x 10) + (4 x 50) = 220

1. CONCEPT MOL
1 mole is a unit of chemical number whose number of atoms or molecules is equal to the Avogadro number and its mass = Mr compound.
If the number Avogadro = L then:
L = 6.023 x 1023
1 mol atom = L of atomic fruit, its mass = Ar of the atom.
1 mol molecule = L molecular fruit mass = Mr molecule tersberut.
The mass of 1 mol of substance is called the molar mass of the substance
Example:
How many molecules are there in 20 grams of NaOH?
Answer:
Mr NaOH = 23 + 16 + 1 = 40
Mole NaOH = mass / Mr = 20/40 = 0.5 mol
The number of molecules NaOH = 0.5 L = 0.5 x 6.023 x 1023 = 3.01 x 1023 molecule.

H. EQUATION REACTIONS
EQUAL REACTIONS HAVE THE PROPERTY
1. The types of elements before and after the reaction are always the same
2. The number of each atom before and after the reaction is always the same
3. The comparison of the reaction coefficient denotes the mole ratio (specifically the gaseous coefficient ratio also denotes the volume ratio provided the temperature den pressure is the same)
Example: Find the reaction coefficient of
HNO3 (aq) + H2S (g) ® NO (g) + S (s) + H2O (l)
The easiest way to determine the coefficient of reaction is to assume the coefficients of each a, b, c, d and e so that:
A HNO3 + b H2S ® c NO + d S + e H2O
Based on the above reaction then
Atom N: a = c (before and after reaction)
Atom O: 3a = c + e ®  3a = a + e ®  e = 2a
Atom H: a + 2b = 2e = 2 (2a) = 4a ® 2b = 3a ®  b = 3/2 a
Atom S: b = d = 3/2 a
So in order to solve we take any price eg a = 2 means: b = d = 3, and e = 4 so the equation of the reaction:
 2 HNO3 + 3 H2S ® 2 NO + 3 S + 4 H2O

22 komentar:

  1. Whais the benefit we studying of stoiciometry?

    BalasHapus
    Balasan
    1. The benefit of studying stoichiometry is to know more about the results of reactants, and to know the law of law in stoichiometry, and to determine the mass, the volume of a compound

      Hapus
  2. Why legal comperative law = legal dalton?

    BalasHapus
    Balasan
    1. Since this Law equals a comparison if two elements react to form two or more compounds, the weight ratio of one element reacting with a certain weight of the other element in both compounds is always a simple integer ratio. [1] For example carbon reacts with oxygen to form carbon dioxide (CO2) and carbon monoxide (CO)

      Hapus
  3. What different mass atom with mass formula?

    BalasHapus
    Balasan
    1. The atomic mass (ma) of a chemical element is the mass of an atom at rest, generally expressed in terms of atomic mass, whereas the mass formula is a principal quantity with a kilogram unit.

      Hapus
  4. Give an example of the problem with the settlement using the boyle law

    BalasHapus
    Balasan
    1. Suppose the initial volume and gas pressure of 3L and 4 atm respectively, then using Boyle's law, the final gas pressure can be calculated if the volume decreases to 2.5 L. According to Boyle's law, P1V1 = P2V2
      Thus, P2 = P1V1 / V2
      => P2 = (3 * 4) / 2.5 = 4.8 atm.
      Therefore the final gas pressure is 4.8 atm.

      Hapus
  5. What is the relationship of mole to mass?

    BalasHapus
    Balasan
    1. The molar mass (mm) represents the mass possessed by 1 mol of the substance. The mass of 1 mole of the substance is equal to the relative molecular mass (M) of the substance in units of gram / mol.
      The relationship of the number of moles (n) to the mass of the substance (m) by the manifestation:
      The formula determines the relationship of the number of moles (n) to the mass of the substance (m), namely:
      M = n × mm or m = n × Ar or m = n × Mr.

      Information :
      M = mass of substance (gram)
      N = number of moles (mole)
      Mm = molar mass = Ar or Mr (gram / mol)

      So to look for many moles (n):
      N = mAr or n = mMr

      Hapus
  6. How to determine stoichiometric point??

    BalasHapus
    Balasan
    1. The maximum point obtained in the 0.1 M NaOH and 1 M HCl systems was mixing between 25 ml of NaOH and 5 ml of HCl (0.5,1.5) and the minimum point of mixing 10 ml of NaOH and 20 ml of HCl (0.05.0.25) The maximum point is reached at mixing the same volume solution

      Hapus
  7. Mention and give examples of the difference between boyle law and dalton law?

    BalasHapus
    Balasan
    1. Boyle's Law, the law of physics that explains how the relationship between the pressure and volume of a gas and Boyle's Law, namely the laws of physics that explain how the relationship between pressure and volume of a gas.
      Example of boyle law:
      A sealed room contains 200 ml volume of gas. If the pressure of the room is 60 cmHg, calculate the gas pressure in the room with a volume of 150 ml?

      Given: V1 = 200 mL; P1 = 60 cmHg; V2 = 150 ml

      Asked: P2?

      Answer:

      P_ {1} V_ {1} = P_ {2} V_ {2} \ text {or} P_ {2} = \ frac {P_ {1} V_ {1}} {V_ {2}}

      P_ {2} = \ frac {200 \ text {ml} .60 \ text {cmHg}} {150 \ text {ml}} = 80 \ text {cmHg}
      Example of dalton's law:
      Sulfur (S) and oxygen (O) form two types of compounds. The sulfur content in compound I is 50% and II 40%. Does Dalton's law apply to such compounds?

      Answer:

      To check the validity of Dalton's law, follow the following ways:

      Determine the ratio of the mass of the elements in each of the compounds in which one element is assigned with the same appeal number, ie = 1.
      Determine the ratio of the mass of the other elements in the compounds.

      Hapus
  8. can you give me example for Relative Atomic Mass (Ar) and Relative Molecular Mass (Mr)?

    BalasHapus
    Balasan
    1. Determine Mr. The following compounds if known Ar H = 1, O = 16, S = 32
      A. H2O
      B. H2SO4
      A: a. Mr. H2O = (2 x Ar H) + (1 x Ar O)
      = (2 x 1) + (1 x 16)
      = 18

      B. Mr H2SO4 = (2 x Ar H) + (1 x Ar S) + (4 x Ar O)
      = (2 x 1) + (1 x 32) + (4 x 16)
      = 2 + 32 + 64
      = 98

      Hapus
  9. Describe the molar volume of the state at standard temperature and pressure.

    BalasHapus
    Balasan
    1. «Hydrocarbon classification
      Relation of Mol (n) with Number of Particles (X), and Molar Mass »
      Molar Gas Volume in Standard Conditions and Circumstances

      September 15, 2011 by naneyanar

      The Avogadro hypothesis states that at the same temperature and pressure, all gases of the same volume will contain the same number of particles. Since 1 mole of each gas has the same number of molecules, at the same temperature and pressure, 1 mole of each gas has the same volume. The volume per mole of gas is called the molar volume and is denoted Vm.

      V = n × Vm

      with:

      V = volume of gas (liter)

      N = number of moles (mole)

      Vm = molar volume (liters / mol)

      (Martin S. Silberberg, 2000)

      The molar volume of the gas depends on the temperature and pressure. Some conditions of temperature and pressure are commonly used as a reference for determining the volume of gas as follows.

      1. Standard Condition

      Conditions with 0 ° C and 1 atm pressure are called standard states and represented by STP (Standard Temperature and Pressure).

      PV = nRT

      with:

      P = pressure (atm)

      V = volume of gas (liter)

      N = number of moles (mole)

      R = gas constant = 0.082 L atm / mol K

      T = 0 ° C = 273 ° K

      V = \ tfrac {nRT} {p}
      V = \ tfrac {1 mol \ times 0,082 L atm / mol K times 273K} {1 atm}
      V = 22.4 liters

      Thus, in the standard state (STP), the molar volume (volume 1 mole of gas) is 22.4 liters / mol.

      Hapus
  10. What is the phase of an influential compound in chemical calculations?

    BalasHapus
    Balasan
    1. Effect because phase (g) for gas
      (L) for pure liquid
      (S) for solid (solid)
      (Aq) for the solution (water soluble) and One important aspect of chemical calculation is the quantitative relationship between the substances involved in the chemical reaction, either as a reactant or as a reaction product. Stoichiometry (stoi-kee-ah-met-tree) is a field in chemistry involving quantitative relationships between substances involved in chemical reactions, either as reagents or as a reaction product. Stoichiometry also involves the ratio of atoms between elements in a chemical formula, for example the ratio of the H atoms and the O atoms in the H2O molecule.

      Hapus
  11. if we want to know the mole of product in the reaction,what should we do in the first step?

    BalasHapus
    Balasan
    1. The first thing to do is to equalize the reaction

      Hapus