stoikiometri reaksi
TRANSCRIPT
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PUJI LASTARIA1C312075
KHUSNUL HATIMAHA1C312021
NOVITASARIA1C312030
ALFIA RIANIA1C312080
WENNY REZKY AMELIAA1C312032
MIFTAH FARIDA1C312013
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After learning this chapter, you are expcted to
be able to:
a. Prove based on experiment that substance mass before and after
reaction is constant (law of conservation of mass/Lavoisier’s law).
d. Use experimental data to prove volume ratio law (Gay Lussac’s
Law).
b. Prove in accordance with experiment and interpret data
concerning two elements mass that compounds (Proust’s law).
e. Estimate gases volume of reactants or products based on Gay
Lussac’s law.
c. Prove the applicable of ratio multiplication law (Dalton’s law) on
several compound.
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f. Find gases volume relation with number of molecules
measured at the same temperature and pressure (Avogadro’s
Law).
g. Explain mole definition as a unit amount of substance.
i. Determine emprical formula, molecular formula and water
crystal and percent composition of compounds.
h. Converse total mole with number of particles, mass and
substances volume.
j. Determine limiting reagent in a reaction.
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Fundamental Laws of Chemistry
Chemical equation
Avogadro’s Hypothesis
Mole Concept
Reaction Stoichiometry
Compound of Stoichiometry
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Fundamental Laws of Chemistry
Law of
Conservati
on of Mass
Proust’s
Law
Dalton’s
Law
Gay
Lussac’s
Law
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The Law of Conservation of Mass
“In every chemical transformation, an equal quantity of matter exists before and after the reaction.”
Fundamental Laws of Chemistry
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At 2,4 grams magnesium burning in air, magnesium oxide produce 4 grams. How many grams of oxygen are used up in the reaction?(Ar Mg = 24 Ar O = 16 )
Example:
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The Law of Definite Proportions
“ In a given chemical compound the proportion by mass of the elements that compose it are fixed, independen of the origin of the compound or its mode of
preparation.”
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Example :
Analysis of the salt from various
regions.
Check to see it meets law of proust?
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Solution :
As shown in the calculation above, the mass
ratio of Na to Cl apparently fixed. Namely 1 :
1,54. so fulfill the law of compound proust
Salt of Mass of
sodium
(Na)
Mass of
chlorine
(Cl)
Mass
Na : Cl
Indramayu
Madura
Impor
0,786 g
0,59 g
0,983 g
1,214 g
0,91 g
1,517 g
0,786 g : 1,214 g = 1 : 1,54
0,59 g : 0,91 g = 1 : 1,54
0,983 g : 1,517 g = 1 : 1,54
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The Law of Multiple Proportions
“When two elements form a series of
compounds, the masses of one that combine with a fixed mass of the other are in the ratio of smallintegers yo each other.”
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Example :
Sulfur (S) and oxygen (O) formed two types of compounds. Levels of sulfur in compounds I and II in a row is 50% and 40%. Whether the law applies to the compound dalton.
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Solution :
Compound I consists of 50% sulfur, the mass oxygen is 50%.Compound II consists of 40% sulfur, the mass oxygen is 60%.Mass S : O of compound I = 50 : 50 = 1 : 1Mass S : O of compound II = 40 : 60 = 2 : 3 or 1 : 1,5If mass S in the compound I = compound II, as equally as 1 gram, then mass O of compound I : II = 1 : 1,5 = 2 : 3.Comparison is a simple integer and two compounds hat fulfill the law of dalton.
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The Law of Combining Volumes
“When two gases are allowed to react, such that the goes are at the
same temperature and pressure, the volumes of each gas consumed will be
in the ratio of small integers. Morever, the ratio of the volumes of
each product gasto the volume of either reacting gas will be ratio of
simple integers.”
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Example :
Two researchers independently studied thedecomposition reaction of dinitrogen pentaoksidabe nitrogen dioxsyde and oxygen. The researcherfound that the decompotion of 50mL (100°C, 1atm) nitrogen pentaoksida produce 100mL(100°C, 1 atm) nitrogen dioxyde and 25mL(100°C, 1 atm) oxygen. Whether the result ofthese trials fulfill the law of combining volumes?
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Solution : 50mL N2O5 100mL NO2 + 25mL O2
2 4 1Then
2N2O5 4NO2 + O2
This statement is meet
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Chemical Equations
Lavoisier, 1788
Because of the principle of the law conservation of mass,
an equation must be balanced.
It must have the same number of atomsof the same kind
on both sides.
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Balancing Equations
Depict the kind of reactants and products and their relative amounts in a reaction.
___ Al(s) + ___ Br2(liq) ---> ___ Al2Br6(s)
2 Al(s) + 3 Br2(g) ---> Al2Br6(s)
The numbers in the front are called
stoichiometric coefficients
The letters (s), (g), and (l) are the physical states of compounds.
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Equation is said equivalent if the type and number of atoms reacted substances (reactants) together with the type and number of atoms of the reaction (products)
Reactants are written on the left followed by an arrow and then the product
2H2 (g) + O2 (g) → 2H2O (l)
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AVOGADRO’S HYPOTHESIS
"At the same temperature
and pressure, all gases
with the same volume
will contain the same
number of molecules as well."
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Exercise Problem
A hydrocarbon compounds (C XHY) by burning gaseous
reaction:
C x H y (g) + O 2 (g) → CO 2 (g) + H 2 O (g) (not equal)
Of the experiment is known that 2 liters of gas to burn C x
H y (T, P) required 5 liters of oxygen gas (T, P) and
produced 4 liters of carbon dioxide gas (T, P). Determine
the molecular formula of the hydrocarbon?
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Answer
Due to volume ratio is a reaction coefficient, then theequation becomes:2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + .... H 2 O (g) (notequal)For equality oxygen atom, then the coefficient of H 2 Ois 2 (10-8), thus the equations become:2 C x H y (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O (g)For C and H atoms equality as follows.Thus, the hydrocarbon molecular formula is C 2 H 2.
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Mole Concept
Mole definition
Relation with the
number of moles of particles
Molar mass and
Molar volume
Calculating Mass and volume of Product
Kemolaran Solution
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Mole Definition
In Chemistry, number of atomic particles or elements that involved in chemical reaction explained by mole.
Mole is a unit to express the number of particles.
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Relationship Between Moles and Number of Particles
The number of particles in the 12 gram 12Catom specified based experiment result is
6,02 x 1023
Relationship Between the number of moles (n) and the number of particles (x) can be formulated as follows :
x = n × 6,02 × 1023
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E AX M
PL
E
How many particles are found in 3 mol of Iron metal?
Answer :1 mol of Iron metal (Fe) = 6,02 × 1023
3 mol of Iron metal = 3 × 6,02 × 1023
= 1,806 × 1024
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M
O
AL R
Mass &Volume
Molar mass in one mol mass of substance expressed in grams.
Molar Mass
m = mass (g)m = n × Mm n = number of mole (mol)
Mm = molar mass(g/mol)
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PL
E
Known Ar Fe = 56. Calculate the molar mass of Fe ?
Answer :
Mass of 1 mol of Fe = 1 × 56 = 56 g
Thus, Molar Mass of Fe = 56 g/mol.
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Molar volume expresses volume for each1 mol of gas. The gas molar volume is 22,4L at STP and 24,5 L at room temperature.
Molar Volume
If expressed at STP molar volume is symbolized by Vm . Relationship between
volume, number of moles, and the gas molar volume is as follows
V = Volume of the gas V = n × Vm n = Number of moles
Vm = The gas molar volume
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When volume of the gas is measuredat certain temperature andpressure, then equation used is isideal gas equation.Mathematically, the ideal gasequation is as follows :
P = pressure of the gas (atm)PV = nRT V = volume of the gas (L)
n = number of molesV = nRT R = gas constant (0,0821 atm L mol-1 K-1
P T = absolute temperature (K)
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E AX M
PL
E
At certain temperature and pressure,5 mole of SO2 gas have 100 L volume. At the same temperature
and pressure . What is the volume of 3 mole of No2 gas?
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Comparison of mole of SO2 gas and NO2 gas = 5 : 3
Volume of 5 mole of SO2 gas is 100 L then thevolume of 3 mol of NO2 is
Thus, volume of NO2 gas is 60 L
A SN W
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Calculating Mass and Volume of Product
There following steps can determine mass or volume of product.
a. Calculate the number of moles of elementary unit (atoms, molecules or ions) from elements, compounds, or ions from known mass or volume of subtance.
b. Calculate the number of moles of the unknown subtances using subtance coefficient in balanced equation.
c. Determine mass of volume of the unknown subtances based on the number of moles calculated.
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Calculate mass of carbon dioxide gas (CO2) produced when 108 g ethane (C2H6) burned in O2
gas.
Answer :Chemical equation :
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)
known 108 g of C2H6, Mr C2H6 = 30 and Mr CO2
= 44
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Three following steps can determine mass ofcarbon dioxide :Step I : Calculate moles of 108 g of ethane
1 mol of C2H6 = 30 g then 108 g C2H6 = 108 g = 3,6 mol30 g/mol
Step II : Calculate the number of moles of carbon dioxide produced.
Based on equation : 2 mol C2H6 produced 4 mol CO2. 3,6 mol C2H6 produce 4 x 3,6 mol = 7,2 mol CO2
2
Step III : Calculate the number of carbon dioxide produced in gram,1 mol CO2 = 44 g7,2 mol CO2 = 7,2 x 44 g = 316,8 g
Thus, the mass of carbon dioxide gas produced in thecombustion of 108 g ethane is 316,8 g. Menu
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One way to express concentration of solution used in chemistry is kemolaran (M).
kemolaran expressed in number of moles of solute per liter of solution, or the amount of
solute in mmol per mL of solution.
M = Kemolaran solutionM = n = number of moles of solute
V = volume of solution
nV
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Calculate the number of moles and the mass of urea (Mr = 60) were present in 200 mL of 0.4 M urea.
Answer :determine the mass of solute,n = V x M• number of moles of urea (n) = V x M
= 0,2 L x 0,4 Mol L-1
= 0,08 Mol
mass urea (m) = n x Mm
= 0,08 mol x 60 g mol-1` = 4,8 gram
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Reaction Stoichiometry
Stoichiometry
coefficients of
the reaction as
the basis of the
reaction
Limiting
Reagent
Chemical
Formula of
hydrates
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As the reaction stoichiometry coefficients Elementary Reactions
Reaction coefficient is the ratio of particles of substances involved in the reaction. Therefore 1 mole of any substance contains the same number of particles, the reaction coefficient is also a comparison of the number of moles of substances involved in the reaction.
Number of moles = (coefficient of substances were asked) / (coefficient of known substances) x number of substances known
Reaction stoichiometry
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Aluminum dissolves in sulfuric acid produce aluminum sulfate and hydrogen gas.
2AL (S) + 3H2SO4 (aq) AL2 (SO4) 3 (aq) + 3H2 (g)
How many moles of hydrogen gas can be produced when 0.5 moles of aluminum used
EXAMPLE
ANSWER
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In the know: Coefficient substances were = 3Coefficient of substances known to = 2The number of moles of Al = 0.5 mol
In question: the number of moles of H2 = ...?
Number of moles = (coefficient of substances were) / (coefficient of unknown substances) x number of substances knownThe number of moles of H2 = (coefficient H2) / (coefficient Al) x number of moles of Al
= 3/2 x 0.5 mol= 0.75 mol
ANSWER
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The reactans used up first a rection andrestricts the running reaction so that nofurther one called limiting reagent.
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Methane burns (reacts with oxygen) by the equation:
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)
In an experiment, a total of 8 grams of methane burned with 40 grams of oxygen.(Ar C = 12; H = 1; O = 16)Determine the limiting reagent.
EXAMPLE
ANSWER
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In the know: Mass CH4 = 8 gramsMass O2 = 40 grams
In question: limiting reagent = ...?
The number of moles of methane (CH4) = (8 g) / (16 g / mol) = 0.5 molThe number of moles of oxygen (O2) = (40 g) / (32 g / mol) = 1.25 mol
When compared with the coefficient of the reaction, methane is multiplied by the number 0.5 / 1 or 0.5, while the number of oxygen with 1.25 / 2 or 0.625.Because pengalinya smaller, the limiting reagent is methane.
ANSWER
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Chemical formula Hydrant
Hydrants are solids that bind several molecules of water as part of its crystal structure.
British salt, MgSO4, 7H2O: Magnesium sulphate pentahydrate
If a hydrate in the heat, some or all of the water crystals can be separated (yawn).
CuSO4. 5H2O (S) CuSO4. (S) + 5H2O (g)If a hydrate is dissolved in water, it will loose its crystalline water.
CuSO4. 5H2O (S) CuSO4. (Aq) + 5H2O (l)
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A total of 10 grams of hydrated iron (II)sulphate crystals are heated so that allwater is evaporated. The remaining solidmass was 5.47 grams. How is the formulathat hydrates?
(Ar H = 1; O = 16 S = 32; Fe = 56)
EXAMPLE
ANSWER
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Differences existing mass is the mass of the crystal water.For example the amount of water crystals is x, so it is FeSO4.xH2O formula hydrates.
FeSO4.xH2O mass = 10 gramsFeSO4 mass = 5.47 grams
Mass of water = 10 to 5.47 = 4.53 grams.
ANSWER
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FeSO4.xH2O (s) FeSO4 (aq) + xH2O (g)
The number of moles of FeSO4 = (5.47 g) / (152 g / mol) = 0.036 molThe number of moles of H2O = (4.53 g) / (18 g / mol) = 0.252 molFeSO4 mol: mol of water = 0.036: 0.252 = 1: 7Meaning, one molecule binds FeSO4 7 water molecules.The formula hydrates is FeSO4.7H2O.
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Compound of Stoichiometry
Composition
Percentage
Empirical
and
Molecular
Formulas
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Composition percent is determined bydividing mass of each element in one molof compound with molar mass ofcompound and multiply by 100 percent.
Composition Percentage
Compound stoichiometry
Example
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Determine percentage of each element in ether anaesthesia, C4H10O. (Ar C = 12, H = 1, O = 16).
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Empirical and Molecular Formulas
A molecular formula shows the exactnumber of atoms of each element in thesmallest unit of a substance.
Empirical formula is the simplest ratioof atoms that compose a molecule, hence,empirical formula is also called comparisonformula.
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A compound consists of 84 % carbon and 16 % hydrogen. If Ar C = 12, Ar H = 1, and Mr =
100, determine the empirical and molecular formulas of the compound.
Answer:
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