STOICHIOMETRY

1. OBJECTIVEa. Examine the truth of the law of conservation of massb. Find the relationship of mass of the elements in compoundsc. Determine the mol ratio of reacted substances

2. BASIC THEORYChemical stoichiometry deals with mass relations in chemical reactions.

Stoichiometry is based upon the fact that mass is conserved in chemical reaction. This means that total mass of the products in a reactions equal the total mass of the reactants used up in the reactions. (Dickson. Laboratory Experiments for Introduction Chemistry, 115). The French chemist, Antoine Lavoiser found that mass is conserved in a chemical reaction. The total mass of the products of a chemical reaction is always the same as the total mass of the starting materials consumed in the reaction. His results led to one of the fundamental laws of chemical behavior: the law of conservation of matter, which states that matter is conserved in a chemical reaction. (http://chemed.chem.purdue.edu/genchem/history/lavoisier.html)

Precipitation reaction define as one in which an insoluble solid (the precipitate) forms when two or more aqueous solutions are mixed. If you combine these two solution, all of the cations and anions suddenly find themselves in the same solution.(Russo, Steve and Silver, Make. Introductory Chemistry, 256 and 257) In chemical reaction, reactants will react and form products. This way the amount of reactants decrase while the amount of products increases. The change in concentration of reactants or products per unit time is known as the rate of reaction. (Johari, J.M.C and Rachmawati M. Chemistry 2, 86)The Mole

As the size or mass of an atom is too small to measure, it is impractical to state the amount of a substance. Therefore, a much larger unit is required; the unit is called mole. Mole relates the number of particles in a substance that is stated in mol unit with its mass or volume.

The mole states the number of particles (atoms, molecules, or ions) ina substance, expressed in mol. One mole ( 1 mol ) contains the same number of particles as a 12.0 gram of caebon-12, which is 6.02 x 1023 particles. This value is known as Avogadro’s number (NA), or in Germany Loschmidt number (L). (172) (Johari,J.M.C, and Rachmawati,M.2009.Chemistry 1 for Senior High School Grade X.Jakarta:Erlangga.Determining Molar Ratios

The molar ratios identify how many moles of product are formed from a certain amount of reactant, as well as the number of moles of a reactant needed to completely react with a certain amount of another reactant.

The coefficients in a balanced equation can be used to write a molar ratio. Molar ratios are conversion factors that can be used to relate:

1.) moles of product formed from a certain number of moles of reactant2.) moles of reactant needed to form a certain number of moles of a product.

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3.) the number of moles of a particular reactant needed to completely react with a certain number of moles of a second reactant.(http://www.occc.edu/kmbailey/chem1115tutorials/Molar_Ratios.htm)

Law of Definite Proportion DefinitionIn chemical science, the law of definite proportions, often named Proust's Law

states that, "A chemical compound always holds exactly the same ratio of chemicals

element by mass. The equivalent statement is the law of invariant composition, which in turn states that all samples of a given chemical compound have the same elementary constitution." More clearly, law of definite proportion, one of the laws of chemical composition can be defined as, Every compound always consists of the same elements present together in a fixed proportion by mass.

The laws of chemical combination played an important role in the formulation of atomic molecular theory. These laws summarize the experimental results that are obtained about the masses of the reactants and products in a very large number of chemical reactions.

The various laws are,1. Law of conservation of mass2. Law of definite proportions3. Law of multiple proportionsLaw of Conservation of Mass :This law states that the total mass of the reactants is equal to the total mass of

the products in every chemical reaction. In other words, a chemical reaction is merely a transformation of material substances, it involves neither creation nor destruction of mass.

Law of Multiple Proportions :This law states that whenever two elements combine to form two or more

compounds, the mass of one of the elements that combine with a fixed mass of the other element is in a ratio of small integral numbers.Or when two elements combine to form two or more compounds, the ratio in which one of the elements combines with a fixed mass of the other is always simple. (http://chemistry.tutorvista.com/inorganic-chemistry/law-of-multiple-proportions.html)

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Weight the erlenmayer and its and record the mass

3. APPARATUS AND REAGENTa. Erlenmayer flask 80 ml (4) a. NaOH 2 M solutionb. Erlenmayer flask 250 ml (1) b. H2SO4 2 M solutionc. Test tube 75x 12 mm (10) c. CuSO4 0,1 M solutiond. Analytical Balance d. KI 0,1 M solutione. Burner e. Pb(NO3)2 0,1 M solutionf. Stative and Clamps f. Copper metal (plate 30x 4 mm)g. Spatula g. Sulphur Powderh. Graduated cylindersi. Beaker j. Thermometer

4. PROCEDUREa. The Conservation of mass in chemical reactions

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10 ml of NaOH 5 ml of CuSO4 Insert the test tube into the erlenmayer

Tilt the erlenmayer until the two solutions can be mixed

Weight the erlenmayer and its contents again

Filter the solution and record the result.

Conduct the same experiments using 10 ml of KI solution and 5 ml of Pb(NO3)2 solution.

b. The relationship between the mass elements in compound

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Taking one spatula tip of sulfur powder and is put into metal spoon, then a copper plate is added.

Holding the metal spoon and heating it slowly until sulfur used up.

Measuring the length of remaining copper.

In the same way, heating the other copper plates and more amount of sulfur, which is 2, 3, 4, and 5 times the first one.

Drawing a graph between the length of the remaining copper vs the amount of used sulfur.

c. Determination of reactant moles ratio

d. Acid-base Stoichiometry

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0,1 M Pb(NO3)2

0,1 M KI

A B C D E F

Put 0,1 M Pb(NO3)2 solution and 0,1 M KI solution into six test tubes according to the table

A

B

C

D

E

F

Stir each mixture, then let the sediment go down completely

Measure te height of the precipite

Test Tube A B C D E FVolume Pb(NO3)2 (ml) 1 2 3 5 7 9Volume KI (ml) 9 8 7 5 3 1

5. OBSERVATION SHEETa. The conservation of mass in chemical reactions

1.) Reaction between NaOH and CuSO4 solution Erlenmayer mass before reactions : 134,5767 gram

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5 ml 10 ml 15 ml 20 m 25 mlNaOH

5ml 10ml 15ml 20ml 25ml H2SO4

Measure and record the temperature of each solution then take the average temperature

The solution are mixed in such away that the volume of each mixture is 30 ml

Observe the temperature change during the mixing and record the final temperature as. Calculate the temperature difference.

Erlenmayer mass after reactions : 135,5734 gramThe change of substance states after filtration

Before reaction after reactionNaOH (aq) + CuSO4 (aq) Na2SO4(aq) + Cu(OH)2 (s)

transparent blue transparent Blue2.) Reaction between KI and Pb(NO3)2

Erlenmayer mass before reactions : 135,1010 gramErlenmayer mass after reactions : 135,0979 gramThe change of substance states after filtration

Before reaction after reaction2KI (aq) + Pb(NO3)2 (aq) 2KNO3 (aq) + PbI2 (S)

Transparent transparent transparent yellowb. The relationship between the mass elements in the compound

Amount of(spatula tip)

1 2 3 4 5

The length ofremained Cu (mm)

12 11,5 12 11,8 12

Reaction :Cu (s) + S (s) CuS (g)

1 2 3 4 511.2

11.3

11.4

11.5

11.6

11.7

11.8

11.9

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12.1

Chart Title

The lenght of remained Cu (mm)

Amount of S

the

leng

ht o

f rem

aine

d Cu

Figure 1 . The relationship of the lenght of remained Cu vs the amount of S

c. Determine of reactant moles ratio

Test tube 1 2 3 4 5 6Vol Pb(NO3)2 (ml) 1 2 3 5 7 9Vol KI (ml) 9 8 7 5 3 1

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Height of precipite (mm) 4 5,5 6 6 5,5 5

The height of precipitate (mm)

9 8 7 5 3 11 2 3 5 7 9

0

1

2

3

4

5

6

7

V Pb(NO3)2 (mL)V KI (mL)

Hig

h of

pre

sipi

tate

(mm

)

Figure 2. The relationship of height of precipitate vs volume Pb(NO3)2 or volume KI

The reaction

Pb(NO3)2 + 2KI PbI2 + 2KNO3

d. Acid Base Stoichiometry

Volume of NaOH (ml)

TM (NaOH)0C

Volume of H2SO4 (ml)

TM (H2SO4)0C

TM Ti ΔT

0 - 30 30,5 30,5 30,5 05 29 25 29 29 30 110 29 20 30 29,5 30 0,515 30 15 30 30 31 120 30 10 30 30 31 125 30 5 32 31 29,5 0,530 29,5 0 29,5 29,5 29,5 0

Reaction2NaOH(aq) + H2SO4 (aq) Na2SO4(aq) + 2H2O (l)

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30 25 20 15 10 5 00 5 10 15 20 25 30

0

0.2

0.4

0.6

0.8

1

1.2Relationship ΔT vs volume of acid-base

V NaOH (mL)V H2SO4 (ml)

ΔT

Figure 3 Relationship ΔT vs volume acid-base

6. DATA ANALYSISa) The conservation of mass in chemical reactions

NaOH (aq)+CuSO4(aq )→ Cu (OH )2(aq)+Na2 SO4 (s)

134, 5767 gram 134,5739 gram

2 KI (aq)+Pb(NO¿¿3)2 (aq )→ PbI2 (s)+2 KNO3(aq)¿

135,1010 gram 135, 0986 gram

a) The relationship between the mass elements in the compoundAmount of sulfur (spatula tip) 1 2 3 4 5

The length of remained Cu (mm) 12 11,5 12 11,8 12

Δ length (mm) 0 0,5 0 0,2 0

b) Determination of reactant moles ratio1) Reaction in the first test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,1 mmol 0,9 mmolR 0,1 mmol 0,2 mmol 0,1 mmol 0,2 mmolS - 0,7 mmol 0,1 mmol 0,2 mmol

Mass of PbI2=0,1 X 461=46,1 mg

2) Reaction in the second test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,2 mmol 0,8 mmol

R 0,2 mmol 0,4 mmol 0,2 mmol 0,4 mmol

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S - 0,4 mmol 0,2 mmol 0,4 mmol

Mass of PbI2=0,2 X 461=92,2 mg

1) Reaction in the third test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,3 mmol 0,7 mmol

R 0,3 mmol 0,6 mmol 0,3 mmol 0,6 mmol

S - 0,1 mmol 0,3 mmol 0,6 mmol

Mass of PbI2=0,3 X 461=138,3 mg

2) Reaction in the forth test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,5 mmol 0,5 mmol

R 0,25 mmol 0,5 mmol 0,25 mmol 0,5 mmol

S 0,25 mmol - 0,25 mmol 0,5 mmol

Mass of PbI2=0,25 X 461=115,25 mg

3) Reaction in the fifth test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,7 mmol 0,3 mmol

R 0,15 mmol 0,3 mmol 0,15 mmol 0,3 mmol

S 0,55 mmol - 0,15 mmol 0,3 mmol

Mass of PbI2=0,15 X 461=69,15 mg

4) Reaction in the sixth test tube

Pb(NO¿¿3)2 (aq)+2KI (aq )→ PbI2 (s)+2 KNO3(aq)¿

M 0,9 mmol 0,1 mmol

R 0,05 mmol 0,1 mmol 0,05 mmol 0,1 mmol

S 0,85 mmol - 0,05 mmol 0,1 mmol

Mass of PbI2=0,05 X 461=23,05 mg

c) Acid-base Stoichiometry1) Reaction between 0 ml NaOH and 30 mL H2SO4 (ΔT = 0)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x0 ¿0,2 x30 ¿0 mmol ¿6 mmol

The ratio between NaOH and H2SO4 = 0 : 6

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2) Reaction between 5 ml NaOH and 25 mL H2SO4 (ΔT = 1)mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x5 ¿0,2 x25 ¿1 mmol ¿5 mmol

The ratio between NaOH and H2SO4 = 1 : 53) Reaction between 10 ml NaOH and 20 mL H2SO4 (ΔT = 0,5)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x10 ¿0,2 x20 ¿2 mmol ¿4 mmol

The ratio between NaOH and H2SO4 = 1 : 24) Reaction between 15 ml NaOH and 15 mL H2SO4 (ΔT = 1)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x15 ¿0,2 x15 ¿3 mmol ¿3 mmol

The ratio between NaOH and H2SO4 = 1 : 15) Reaction between 20 ml NaOH and 10 mL H2SO4 (ΔT = 1)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x20 ¿0,2 x10

¿4 mmol ¿2 mmol

The ratio between NaOH and H2SO4 = 2 : 16) Reaction between 25 ml NaOH and 5 mL H2SO4 (ΔT = 0,5)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x25 ¿0,2 x5

¿5 mmol ¿1 mmol

The ratio between NaOH and H2SO4 = 5 : 17) Reaction between 30 ml NaOH and 0 mL H2SO4 (ΔT = 0)

mol NaOH=M x V mol H 2 SO4=M xV

¿0,2 x30 ¿0,2 x0

¿6 mmol ¿0 mmol

The ratio between NaOH and H2SO4 = 6 : 0

7. DISCUSSIONThe stoichiometry experiments have been done by treating different

experiment can be described as below:a. The conservation of mass in chemical reactions

The first experiment is seen that after the solution was mixed with each other and both react marked with color change occurs,  in a mixture of NaOH and CuSO4 is blue color changes (precipite)

Before reaction After reaction

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2NaOH (aq) + CuSO4 (aq) Na2SO4(aq)+ Cu(OH)2(s)

(transparent) (blue) (transparent) (blue)The color change proved the chemical reaction is happened. This experiment

also weight the erlenmayer. The result was :Erlenmayer mass before reactions : 134,5767 gramErlenmayer mass after reactions : 135,5734 gramFollowing the result, the mass of the solution before and after the reaction has similarities. So, it shows the truth about the law of conservation of mass.

While the second experiments between KI and Pb(NO3)2 after mixed visible color change to yellow (precipite). Following the reaction :

Before reaction After reaction2KI(aq) + Pb(NO3)2(aq) 2KNO3(aq) + PbI2(s)

(transparent) (transparent) ( transparent) (yellow)The color change proved the chemical reaction is happened. This experiment

also weight the erlenmayer. The result was : Erlenmayer mass before reactions : 135,1010 gramErlenmayer mass after reactions : 135,0979 gramFollowing the result, the mass of the solution before and after the reaction has similarities. So, it shows the truth about the law of conservation of mass.

But the mass before and after in the second experiments has a little difference. A little difference because of the mixed of that solution was not mixed completely. So, it can influent the mass before and after of that solution. But, all of it had can show the truth of the law of conservation of mass by Antoine Lavoiser “The total mass of the products of a chemical reaction is always the same as the total mass of the starting materials consumed in the reaction”.

b. The relationship between the mass elements in compoundThis experiment to prove the Proust Law. In this experiment, we were heating

copper and sulfur to determine the relationship between the mass of the element in the compound. Sulfur initially prepared in a spatula, then add the copper plate and heated until the sulfur depleted. The experiments were performed five times with a growing mass of sulfur.

Long Cu initially on experiment 1,2,3,4, and 5 are 12 mm in length change after reacting to 12 mm; 11,5 mm; 12 mm; 11,8 mm; 12 mm. But, the result was not accurate. The experiment wa not react. Actually, it should be seen that the length of the copper after reacted with sulfur decrease, with increasing sulfur powder reacted the diminishing length of copper. in this case the sulfur acts as a limiting reactant is completely reacted with the copper. the reaction is complete when the sulfur has reacted completely discharged not remained.

Our shaped graph obtained fluctuate (up and down). This is because it is difficult to obtain appropriate results or it is not react completely. We think that there are some factors that might influent the result. Among others in the

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determination of the amount of sulfur is not weighed accurately and the reaction is not completely.

c. Determination of reactant moles ratioIn this experiment, we react KI solution with Pb(NO3)2 solution by six

variation of volume KI and Pb(NO3)2. From these reactions produced the changed of color from transparent to yellow and the variation of the height of precipite.

The data obtained sediment height (PbI2) of the composition of the test tube in the series was 4 mm; 5,5 mm; 6 mm; 6 mm; 5,5 mm; 5 mm. That data is obtained by ratio Pb(NO3)2 and KI. The ratio of Pb(NO3)2 and KI are 1:9 ; 2:8 ; 3:7 ; 5:5 ; 7:3 ; 9:1. From the data obtained sediment height, we know that the highest sediment is 6 mm in the ratio 3:7. It has appropriate with the law of moles ratio. According to the reaction :

2 KI (aq)+Pb(NO¿¿3)2 (aq )→ PbI2 (s)+2 KNO3(aq)¿(Transparent) (Transparent) (Yellow) (Transparent)

After we see the reaction we know that the ratio of coefficient or the ratio of moles we know that the ratio of Pb(NO3)2 and KI is 2:1. So, in the ratio 3:7 is stoichiometric reaction that happened between another reaction, because appropriate with coefficient reaction.

d. Acid-base Stoichiometry

This experiment was used by reacting 1 M NaOH and 1 M H2SO4 by volume of each solution composition varied so as to produce a mixture of 30 ml volume. Data obtained from this experiment with the relationship between the volume of acid-base with ΔT. Comparison of the number of molecules reacting NaOH and H2SO4 is 0 : 30, 1: 5, 1: 2, 1: 1, 2: 1, 5: 1, 30 : 0. According to comparason that molecules, it is obtained ΔT such as 0°, 1°, 0.5°, 1°, 1°,0.5°,0° . From data we know that the highest ΔT is 1° on the comparison 1:5, 1:1 and 2:1.

Before we conclude which is true, we can see the reaction of acid-base from the experiment.

The reaction is : 2NaOH(aq) + H2SO4 (aq) Na2SO4(aq) + 2H2O (l)

According that reaction we know that the comparison of NaOH and H2SO4 is 2:1. So, the highest ΔT is true is on camparison 2:1. Because in that comparison happened exotherm reaction that causes T average between NaOH and H2SO4

lower than T after mixed.We know that exotherm reaction generally the system themperature increase. As for the other is possibility human error. It might be inaccuracy in reading the themperature so obtained the data inaccurate.

8. CONCLUSSIONAccording to the experiment, we conclude that :1.) The law of conservation of mass by Antoine Lavoisier is true. The mass of

substance before and after chemical reaction is same. We can see in the first experiment :a. Reaction between NaOH and CuSO4 solution

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Erlenmeyer mass before reaction : 134,5767 gramErlenmeyer mass after reaction : 134,5734 gram

b. Reaction between KI and Pb(NO3)2

Erlenmeyer mass before reaction : 135,1010 gramErlenmeyer mass after reaction : 135,0979 gram

2.) According to tthe law of Proust and Dalton about multiple proporsion, this experiment conclude that when two elements combine to form two or more compounds, the ratio in which one of the elements combines with a fixed mass of the other is always simple.

3.) The molar ratios are formed from a certain amount of reactant, as well as the number of moles of a reactant needed to completely react with a certain amount of another reactant and its balance with coefficient reaction.

9. SUGGESTION1.) Practitioner should first read and understand the lab manual carefully 2.) Mutual collaboration between group members 3.) Ask the lab assistant or lecturer if there are difficulties 4.) Comply with all rules that exist in the laboratory in order to avoid the things that

are not desirable 5.) Use time as possible 6.) To experiment with copper sulfur should be done outdoors, because the smell of

sulfur was overpowering and it is advisable to wear a mask 7.) For experiments measure the height of the sediment mixture of KI and Pb (NO3)2,

it takes patience to wait until the mixture is completely settles takes carefully and thorough in measuring the sediment.

8.) For acid-base stoichiometry experiment takes precision in reading the thermometer to determine the temperature scale

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10. ATTACHMENT

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11. REFERRENCESDickson, T.R. 1983. Laboratory Experiments for Introduction Chemistry.

United States : Wiley.http://chemed.chem.purdue.edu/genchem/history/lavoisier.html

Russo, Steve and Silver, Make. 2007. Introductory Chemistry. San Fransisco : Pearson Benjamin Cummings.

Johari, J.M.C and Rachmawati M. 2009. Chemistry 2. Jakarta : Erlangga.https://www.boundless.com/chemistry/textbooks/boundless-

chemistrytextbook/mass-relationships-and-chemical-equations-3/reaction-stoichiometry-44/molar-ratios-231-7525/

http://www.occc.edu/kmbailey/chem1115tutorials/Molar_Ratios.htmhttp://chemistry.tutorvista.com/inorganic-chemistry/law-of-multiple-

proportions.html

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12. APPENDIX1. Write down a) reaction equation, b) balance reaction and c) change of state of :

a. Lead nitrate solution reacts with potassium iodide solutionb. Copper sulphate solution reacts with chloride acid solution c. Sodium Carbonate solution reacts with chloride acid solutiond. Sodium acetate solution reacts with Chloride acid solutione. Marble powder solution reacts with chloride acid solution

2. Eleven grams of iron powder is reacted with 6 gram of sulphur powder, according to the equation : Fe(s) + S(s) FeS(s)

a. How many grams of Fe were used to react completely?b. How many grams of S were used to react completely?c. What substance is excessive?d. How many grams of FeS can be produced from the reaction?

3. How many tons of ZA (Zwavelzuur Ammonia) with 90% of ammonium sulphate are formed from the reaction of 2,55 tons of ammonia and 8 tons of sulphuric acid?

Answer :

1. a. Pb(NO3)2 (aq) + 2KI(aq) PbI2 (s) + 2KNO3 (aq) is obtained yellow precipitate of PbI2

b. NaOH(aq) + CuSO4 (aq) Na2SO4(aq) + Cu(OH)2 (s) is obtained blue precipitate of Cu(OH)2

c. Na2CO3(aq) + HCl (aq) 2NaCl (S) + H2CO3 (aq)

d. CH3COONa (s) + HCl (aq) NaCl(s) + CH3COOH(aq)

e. CaCO3(s) + 2HCl(aq) CaCl2(s) + H2CO3(aq)

2. Known : m Fe = 11 gram Ar Fe = 56

m S = 6 gramAr S = 32

Ask : a.) mass of Fe were used to react completely?b.) mass of S were used to react completely?c.) substance is excessive?d.) mass of of FeS can be produced from the reaction?

Answer : Fe(s) + S (s) FeS (s)

M 1156

mol 632

mol

0,1963 mol 0,1875 molR 0,1875 mol 0,1875 mol 0,1875 molS 0,0089 mol - 0,1875 mol

a.) mass of Fe were used to react = 0,1875 x 56 = 10,5 gram

b.) mass of S were used to react = 0,1875 x 32

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= 6 gramc.) substance is excessive = Fe

the remaining mass = 0,0089 x 56 = 0,5 gram

d.) mass of of FeS = 0,1875 x 88 = 16,5 gram

3. Known : % (NH4)2SO4 = 90% m NH3 = 2,55 ton m H2SO4 = 8 tonAsk : mass of ZA?Answer : 2 NH3 + H2SO4 (NH4)2SO4

M2,5517

mol 8

58 mol

0,150 mol 0,081 mol R 0,150 mol 0,075 mol 0,075 mol

S - 0,006 mol 0,075 molMass of (NH4)2SO4 = 0,075 x 132

= 9,9 ton

Mass of ZA = 10090

x 9,9

= 11 ton

13. PROBLEM1. On the mass conservation experiment, is there any chemical phenomenon? If any,

what is it?2. Refer to Figure 1. How does the relationship of mass of copper and the reacted

sulphur3. Refer to Figure 2

a. Indicate the volume of reactant composition that produced the highest precipitate

b. How many moles of Pb(NO3)2 and KI wich produced the highest precipitatec. How many mole ratio of Pb(NO3)2 and KI which produced the highest

precipitate4. Refer to Figure 3

a. How many volume of NaOH and H2SO4 mixture that showed the highest ΔTb. How many moles of NaOH and H2SO4 which produced the highest ΔTc. What is the mole ratio of NaOH and H2SO4 which produced the highest ΔT

Answer

1. There, in the mass determination experiments are substances that change color chemical events that occur in mixing NaOH and CuSO4. Where the first colored transparent and blue after mixed change to blue (precipite), as well as the mixture

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of KI and Pb(NO3)2 premises which the color changes from transparent to yellow (precipite).

According to the law of Antoine Lavoiser and the experiment, there are the mass of substance before and after chemical reaction is relative same. The Law of Conservation of Mass is true with this experiment.

2. There are mistakes in our experiment so that make the shaped graphic is up and down. Whereas it should be getting down. Because the relationship between the mass of copper and sulfur which react mass that is more and more the amount of sulfur that reacted sulfur was diminishing residual (reduced length) or otherwise.

3. a. Precipitation is highest in the composition whereby the volume of KI is 7 ml and the volume of Pb(NO3)2 is 3 mlb. mol Pb(NO3)2 = M x Vol. Pb(NO3)2

= 0,1 x 3 = 0,3 mol

mol KI = M x Vol. KI = 0,1 x 7 = 0,7 mol

c.mole ratio Pb(NO3)2 and KI is (according to the mole ratio) :mole Pb(NO3)2 : mole KI Ratio

0,1 : 0,9 1:90,2 : 0,8 1:40,3 : 0,7 1:20,5 : 0,5 1:10,7 : 0,3 2:10,9 : 0,1 9:1

By ratio of coefficients

2KI(aq) + Pb(NO3)2(aq) 2KNO3(aq) + PbI2(s)

So, ratio Pb(NO3)2 : KI which produced the highest precipitate is 1:2. It same with our experiment.

4. a. volume mixture of NaOH and H2SO4 which has the highest price is 20 ml NaOH with 10 ml of . H2SO4

b. mol H2SO4 = M x Vol. H2SO4

= 2 x 10 = 20 mol

mol NaOH = M x Vol. NaOH = 2x 20

= 40 molc.Ratio of NaOH and H2SO4 :

H2SO4 + 2 NaOH Na2SO4 + 2H2O

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So, ratio NaOH : H2SO4 which produced the highest ΔT is 2:1. It same with our experiment.

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