stoichiometry/Quantitative chemistry limiting reactant percentage yield volume of gas concentration of solution and titration

2. 4.STOICHIOMETRY
4.5 limiting reactat
4.6 percentage yield
4.7 Gas volume
4.8 concentration of solution
4,9 Titration

3. 4.5 LIMITING REACTANT
• WHEN ONE OF THE REACTANTS IS IN EXCESS, THE
OTHER REACTANT IS A LIMITING REACTANT THAT IS
COMPLETELY USED UP.
• THIS IS BECAUSE IT IS THE AMOUNT OF THIS
SUBSTANCE THAT DETERMINES THE AMOUNT OF
PRODUCT FORMED IN A REACTION,IN OTHER
WORDS IT IS LIMITING
• FOR EXAMPLE, WHEN MANY FUELS ARE BURNED AN
EXCESS OF OXYGEN IS USED.
• FUELS ARE EXPENSIVE AND IN LIMITED SUPPLY. THE
OXYGEN IS READILY
• AVAILABLE FROM THE AIR AND USING AN EXCESS
OF OXYGEN ENSURES THAT ALL
• THE FUEL BURNS
• MAGNESIUM REACTS WITH SULFURIC ACID ,5 MOLES OF
MAGNESIUM (MG) IS REACTED WITH 7 MOLES OF
SULFURIC ACID (H2SO4). ONE OF THE REAGENTS IS IN
EXCESS.CALCULATE THE MOLES OF THE PRODUCTS
FORMED.
• MG + H2SO4 MGSO4 + H2
ANSWER
AS MAGNESIUM AND SULFURIC ACID REACT WITH THE RATIO
1 TO 1,
ONLY 5 MOLES OF THE SULFURIC ACID CAN REACT AS THERE
ARE ONLY
5 MOLES OF MAGNESIUM TO REACT WITH. THE REST OF THE
SULFURIC
ACID IS IN EXCESS (2 MOLES). THEREFORE, THE MAGNESIUM
IS THE
LIMITING REACTANT AND DETERMINES HOW MUCH OF THE
PRODUCTS
ARE MADE. IN THIS CASE, 5 MOLES OF MGSO4 AND 5
MOLES OF H2
WOULD BE MADE.

4. 4.6 PERCENTAGE YIELD
• PERCENTAGE YIELD IS A MEASURE OF THE AMOUNT PRODUCED IN A REACTION
• COMPARED TO THE MAXIMUM THEORETICAL AMOUNT THAT IS EXPECTED AS A PERCENTAGE. FOR EXAMPLE, IF A
REACTION WAS EXPECTED TO FORM A MAXIMUM THEORETICAL 20 G OF PRODUCT BUT ONLY 10 G WAS MADE,
THEN THE YIELD IS 10 G AND THE PERCENTAGE YIELD IS 50%.
• PERCENTAGE YIELD = 100 × MASS OF PRODUCT ACTUALLY MADE/MAXIMUM THEORETICAL MASS OF PRODUCT
WHEN CARRYING OUT CHEMICAL REACTIONS WE ARE UNLIKELY TO PRODUCE ALL THAT WE EXPECT TO. THERE ARE
MANY REASONS FOR THIS.
1 SOME REACTIONS DO NOT GO TO COMPLETION (I.E. THEY DO NOT COMPLETELYFINISH) – SOMETIMES THIS IS
BECAUSE THEY ARE REVERSIBLE AND SOME OF
THE PRODUCTS MAY TURN BACK INTO REACTANTS.
2 SOME OF THE PRODUCT MAY BE LOST WHEN IT IS SEPARATED FROM THE REACTION MIXTURE – FOR
EXAMPLE, SOME MAY BE LEFT ON THE APPARATUS.
3 SOME OF THE REACTANTS MAY REACT IN WAYS DIFFERENT TO THE DESIREDREACTION – IN OTHER WORDS
SOME OF THE REACTANTS MAY TAKE PART IN
OTHER REACTIONS AS WELL.

5. 4.6 PERCENTAGE YIELD
• EXAMPLE
• IN A REACTION WHERE THE MAXIMUM THEORETICAL MASS OF PRODUCT WAS 40 G, THE YIELD
• PRODUCED WAS 15 G. CALCULATE THE PERCENTAGE YIELD.
• ANSWER
• PERCENTAGE YIELD = 100 × MASS OF PRODUCT ACTUALLY MADE/
• MAXIMUM THEORETICAL MASS OF PRODUCT
• = 100 × 15/
• 40
• = 37.5%

6. 4.7 GASE VOLUME
● THE VOLUME OF GASES
THE VOLUME OF A GAS VARIES WITH TEMPERATURE AND PRESSURE:
● THE HIGHER THE TEMPERATURE OF A GAS, THE GREATER ITS VOLUME.
● THE GREATER THE PRESSURE OF A GAS, THE SMALLER ITS VOLUME.
HOWEVER, PROVIDING THE TEMPERATURE AND PRESSURE OF GASES ARE THE SAME,
EQUAL NUMBERS OF MOLES OF ALL GASES HAVE THE SAME VOLUME
volume (dm3) of one mole of any gass= 24dm
At room temp 20c and pressure 1 atm
1 mole O2 24dm3
1 mole CO2 24dm3
1 mole Ar 24dm3
10 moles of O2: volume = 24 × moles = 24 × 1
dm3
0.2 moles of CH4: volume = 24 × moles = 24 ×
dm3

7. Masses can be converted to moles using the equation
mass = Mr × moles.
This can be used to find the volume of a gas from its mass
and vice versa
Example
What is the volume of 64 g of methane gas at room
temperature and pressure?
Answere
64 g moles of CH4: moles = mass/Mr = 64/16 = 4 moles
volume = 24 × moles = 24 × 4 = 96 dm3
Example
What is the mass of 1.8 dm3 of nitrogen gas measured at
room temperature and pressure?
Answer
1.8 dm3 of N2 = volume/moles
= 1.8 /24
= 0.075 moles
mass = Mr × moles = 28 × 0.075 = 2.1 g

8. ● Reacting volumes of gases
Due to equal amounts of moles of different gases having
the same
volume (at the same temperature and pressure), we can
work out
the volumes of gases involved in chemical reactions.
N2(g) + 3H2(g) 2NH3(g)
1
mole of 3 moles of H2 gas 2 moles of NH3 gas
N2 gas
24 dm3 72 dm make 3 48 dm3
Example
What volume of oxygen reacts with 10 dm3 of hydrogen
with the volume of both
gases measured at the same temperature and pressure?
Answer
2H2(g) + O2(g) 2H2O(l)
2 moles of H2(g) reacts with 1 mole of O2(g)
therefore volume of O2(g) = 1/2 × volume of H2(g) = 1/2
× 10 = 5 dm3

9. 4.8 CONCENTRATION OF SOLUTION G/DM3
• WE CAN MEASURE THE CONCENTRATION OF A SOLUTION BY CONSIDERING WHAT MASS OF SOLUTE IS
DISSOLVED IN THE SOLUTION.THIS IS USUALLY FOUND IN G/DM3, WHICH MEANS THE NUMBER OF
GRAMS OF SOLUTE DISSOLVED IN EACH DM3 OF SOLUTION: 1 DM3 IS THE SAME VOLUME AS 1000 CM3
OR 1 LITRE. FOR EXAMPLE,
• IF 50 GRAMS OF COPPER SULFATE IS DISSOLVED IN 2 DM3 OF SOLUTION, THEN THE CONCENTRATION IS
25 G/DM3
• EQUATION YOU NEED TO USE.
• CONCENTRATION(G/DM3) = MASS G/VOLUME
• IN LABOROTORIES CM3
• 1000CM3=1 DM3 WE SHOULD DIVIDE VOLUME IN CM3 BU 1000 TO GET ANSWERE IN DM3
• 25CM3 = 25/1000 0.025DM3

10. CONCENTRATION OF SOLUTIONS IN MOL/DM3
WE CAN MEASURE HOW CONCENTRATED A SOLUTION IS
IN MOL/DM3. THIS IS
EFFECTIVELY THE NUMBER OF MOLES OF SOLUTE
DISSOLVED IN EACH 1 DM3 OF
SOLUTION ; 1 DM3 IS THE SAME VOLUME AS 1 LITRE OR
1000 CM3.
IF THERE WERE 6 MOLES OF SOLUTE DISSOLVED IN 2 DM3
OF SOLUTION, THEN THE
CONCENTRATION WOULD BE 3 MOL/DM3.
concentration (mol/dm3) =volume (dm3)/moles

11. ●4.9 Titrations
Titrations are a very accurate experimental technique that
can be
used to find the concentration of a solution by reacting it
with a
solution of known concentration. Titrations are often used
to find the
concentration of acids or alkalis.
Titrations use apparatus including a pipette, conical flask
and burette
. A pipette is a glass tube designed to measure a specific
volume of a solution very accurately. A typical pipette
measures out
25 cm3 within a margin of ±0.06 cm3. The pipette is filled
using a
pipette filler which is attached to the end of the pipette. A
burette is
a glass tube with a tap (to let out the liquid) with markings
on to show
the volume to the nearest 0.1 cm3.

12. The following steps are followed in a titration.
1 A known volume of a solution of an acid or alkali is
measured out
using a pipette and placed into a conical flask.
2 A few drops of a suitable indicator are added. For most
acid-alkali
titrations, methyl orange or phenolphthalein is suitable.
3 The other solution, the acid or alkali, is added to the
conical flask
from a burette.
4 The solution is added from the burette until the
indicator changes
colour (the end point). The solution is added dropwise
around the
point where the indicator changes colour to ensure the
exact volume
required is used.
5 The volume added from the burette is recorded.
6 The experiment is repeated until concordant results are
achieved (i.e.

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