laboratory chemistry emergency plan days 6 -...

Laboratory Chemistry Emergency Plan Days 6 - 10 Day 4:

• Skill Building Topic 9 – “A Quantitative Understanding of Formulas”

Day 5:

• Skill Building Topic 10 – “Understanding Mass Relationships in Chemical Reactions”

• Putting It all Together – “You Are the Chemist” -

The Solvay Process Reference: Periodic Table

DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS

Skill Building Topic 9 A QUANTITATIVE UNDERSTANDING OF FORMULAS

Calculating Percent Composition from a Formula The percent by mass of each material found in an item is called the element’s percent composition. To find the percent composition of an element in a compound, first calculate the compound’s molar mass. Then divide the total mass of the element in the compound by the molar mass of the compound and multiply the result by 100%. Example

Calculate the molar mass of sucrose, C12H22O11.

12 mole of C (12.00g/1mole) = 144.0 gram

22 mole of H (1.01g/mole) = 22.0 grams

11 mole of O (16.00g/mole) = 176.0 grams

Molar mass of C12H22O11 = 342.2 grams

Find the mass % of each element in sucrose

C %08.42%100lsucrose/mo g 342.2

C/mol g 144.0100 lsucrose/mo mass

C/mol mass C % =×=×=

O 51.43% 100% lsucrose/mo g 342.2

O/mol g 176.0 100x lsucrose/mo mass

O/mol mass O %

H6.48% 100% lsucrose/mo g 342.2

gH/mol 22.18 100 lsucrose/mo mass

H/molmass H %

=×==

=×=×=

Calculating Percent Composition from Mass In the laboratory, it is possible to determine the percentage composition for many compounds. Example

In the laboratory, 0.400 g magnesium ribbon is burned in the presence of oxygen to form magnesium oxide. The product has a mass of 0.663 g. What is the percentage composition of the compound? Mass of Magnesium: 0.400 g Mass of magnesium oxide: 0.663 g

O 39.7%%3.60%100O %

Mg 60.3% 100compound g 0.663

Mg g 0.400Mg %

=−=

=×=


Formulas from Percentage Composition: Empirical Formulas An empirical formula gives the relative numbers of different elements in a substance, using the smallest whole numbers for subscripts. The empirical formula of the compound can be calculated from percent composition data or from the number of grams of each element in the sample. A formula compares the relative numbers of moles of each element. If percents are used, they must be converted to grams, and grams to moles. It is easiest to assume there is 100 g of the compound. Then the percentages are equal to the number of grams of each element. Example 1

A hydrocarbon (the only elements in the compound are hydrogen and carbon) consists of 85.7% carbon and 14.3% hydrogen. What is its empirical formula? Step 1: Assume that you have 100 g of a compound, and that 85.7 g of it are carbon and 14.3 g are hydrogen. Step 2: Use dimensional analysis to find the number of moles of each element in the compound.

H mol 14.2H g 1.01

H mol 1.00H g 14.3

C mol 7.14C g 12.0

C mol 1.00C g 85.7

=×

=×

Step 3: Determine the smallest whole number ratio of moles of elements by dividing all mole values by the smallest.

H mol 21.997.14

H mol 14.2

C mol 17.14

C mol 7.14

≈=

=

The ratio of moles of C to moles of H = 1:2 and the empirical formula for the compound is CH2.

Example 2

The percentage composition of one of the oxides of nitrogen is 74.07% oxygen and 25.93% nitrogen. What is the empirical formula? Step 1. 100 g of compound consists of 74.07 g oxygen and 25.93 g

nitrogen.

Step 2:

N mol 1.85N g 14.01N mol 1.00N g 25.93

O mol 4.62O g 16.00O mol 1.00O g 74.07

=×

=×

Step 3:

N mol 1.01.85

N mol 1.85

O mol 2.51.85

O mol 4.62

=

=


Step 4: This ratio does not consist only of whole numbers, but formulas are not

expressed with decimals: NO2.5. Multiply all the numbers in the ratio by a number that converts the decimal to a whole number. In this case the number is 2, and the empirical formula becomes N2O5.


Skill Building Topic 9 A QUANTITATIVE UNDERSTANDING OF FORMULAS Practice Exercises Complete the following exercises. 1. Find the percent carbon and percent hydrogen in C2H4.

2. Find the percent carbon and the percent hydrogen in C3H6.

3. What do the answers to problems 1 and 2 have in common? Explain your answer.

___________________________________________________________________________

___________________________________________________________________________

___________________________________________________________________________

___________________________________________________________________________

________________________

4. Find the percent of each element in the following compounds. a. MgSO4

b. MgSO4 • 7 H2O

c. (CH3)2SO

d. C5H5N


e. Ca3(PO4)2

5. A compound consists of 11.66 g Fe and 5.01 g O. What is the percent of each element in the compound?

6. A sample of adrenaline, an important hormone, is made up of 0.590 g carbon, 0.071 g

hydrogen, 0.262 g oxygen, and 0.077 g nitrogen. What is the percent of each element in the compound?

7. Copper (II) sulfate commonly appears as a hydrated salt. In the lab, a sample of hydrated salt was heated until all the water was driven off. The original sample had a mass of 24.50 g. The dried sample has a mass of 15.66 g. What is the percent of water in this compound?

8. A compound is 82.3% nitrogen and 17.6% hydrogen. What is the empirical formula?

9. Another compound is 52.2% C, 13.0% H, and 34.8% O. What is the empirical formula?

10. Nicotine is 74.03% C, 8.70% H, and 17.27% N. What is the empirical formula?


11. A hydrocarbon is found to be 82.66% carbon. What is the empirical formula?

12. A sample of a compound contains 0.252 g titanium and 0.748 g chlorine. Determine the empirical formula of this compound.

DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS

Skill Building Topic 10 UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS If the number of atoms is conserved in a chemical reaction, the mass must also be conserved as expected from the Law of Conservation of Mass. In the equation for the formation of water

2 H2(g) + O2(g) → 2 H2O(l) 2 molecules of hydrogen and 1 molecule of oxygen combine to form 2 molecules of water. We could also say that 2 moles of hydrogen react with 1 mole of oxygen to form 2 moles of water. Using the number of grams in a mole of each substance, the mass relationships in the table can be determined. The ratio of moles of hydrogen to moles of oxygen in forming water will be 2:1. If 10 moles of hydrogen are available, 5 moles of oxygen are required.

2 H2(g) + O2(g) → 2 H2O(l) 2 molecules 1 molecule 2 molecules 2 mol 1 mol 2 mol 2 moles x (2.02g/mol) 1 mole x (32.00g/mol) 2 moles x (18.02g/mol) 4.04 g 32.00 g 36.04 g

Solving problems involving the masses of products and or reactants is conveniently accomplished by dimensional analysis. All numerical problems involving chemical reactions begin with a balanced equation.

Example

Find the mass of water produced when 10.0 grams hydrogen react with plenty of oxygen. 2 H2(g) + O2(g) → 2 H2O(l)

Step 1: We know the amount of oxygen is in excess, so we will focus on hydrogen. Find the moles of hydrogen represented by 10.0 g by using the molar mass of H2.

22

22 Hmol 4.95

Hg 2.02 Hmol 1 Hg 10.0 =×

Step 2: Find the number of moles of H2O produced by 4.95 mole H2. From the balanced equation, we know that for every 2 mol H2 we produce 2 mol H2O.

O Hmol95.4 Hmol 2

O Hmol 2 Hmol 4.95 22

22 =×

Step 3: Find the mass of H2O that contains 4.95 mol H2O by using the molar mass of water.

OH g1.98OH mol 1 OH g 18.02OH mol 4.95 2

2

22 =×

Most chemistry students find it is more convenient to set up all three steps in one problem. Make sure that all labels cancel except the g H2O, an appropriate unit to express the mass of H2O as required in the problem.

O Hg1.98O Hmol 1O Hg 18.02

Hmol 2O Hmol 2

Hg 2.02 Hmol 1 Hg 10.0 2

2

2

2

2

2

22 =×××

Molar mass Coefficients Molar Mass


of H2 in equation of H2O

Understanding Limiting Reactants

In Unit 4, Section B of your ChemCom textbook you will find a conceptual atom-counting method for finding the limiting reactant of an equation. Refer to that discussion before you begin this presentation, which presents two additional methods for finding limiting reactants. If a combustion problem states that excess oxygen is available, we need not concern ourselves with the oxygen-to-water ratio. The mass of water produced is predicted from (and limited by) the mass of hydrogen available. Of course, in the laboratory we often deal with specified masses of each reactant, but that requires an enhanced problem-solving method. Suppose a family wants to make several chile rellenos casseroles to serve at a neighborhood party. The recipe lists required ingredients, which are itemized in the left column of the table in Figure 23. The right column represents a survey of pantry and refrigerator contents.

Required Ingredients Available Ingredients Possible Casseroles 1 27-oz can whole green chiles

2 27-oz cans whole green chiles

2

1 pound Monterrey Jack cheese

3 pounds Monterrey Jack Cheese

3

1 pound cheddar cheese 3 pounds cheddar cheese 3 3 eggs 1 dozen eggs 4 3 Tablespoons flour 5 pounds flour Many 5 oz canned evaporated milk 4 cans evaporated milk 4

Figure 23: Required Ingredients Table How many casseroles can be made? Although four casseroles can be made from the available eggs or milk, there are only enough cans of green chiles for two casseroles. In other words, the number of cans of green chiles can be called the limiting factor. After the two casseroles are prepared, cheese, eggs, flour, and milk will remain, but all the green chiles will be used. Therefore, no more casseroles can be made. In this example, the green chiles are the limiting reactant. The limiting reactant is the reactant that is consumed first and limits the amount of product that can be made. The same principle applies in determining the quantity of product that can be produced in a chemical reaction. Let’s take another look at the reaction of hydrogen and oxygen to produce water, then consider what would happen if 2.00 mol hydrogen and 2.00 mol oxygen were available. How many moles of water can be produced? What is the limiting reactant? Which reactant will be in excess and by how much?

2 H2(g) + O2(g) → 2 H2O(l) The balanced chemical equation states that 2.00 mol hydrogen react with 1.00 mol oxygen. When the reaction is complete, 2.00 mol water are produced and 1.00 mol oxygen remains unreacted. This problem is easy to solve by inspection. A more systematic way to solve the problem is to create an SRF table, as shown in Figure 24. The table is composed of the following lines:

Line 1: The balanced chemical equation is listed. Line 2: The Starting number of moles of each substance is listed. This would be what is available, the same as the ingredients for the casseroles. Line 3: The Reacting ratio determined from the coefficients in the balanced equation is multiplied by x, the basic amount of moles that will react. The reactants are being consumed so a minus sign is placed in front of them. The products are increasing so a plus sign is placed in front of them.


Line 4: This line contains what will be left and what will be formed when the reaction is complete. The values of this line are obtained by adding lines 2 and 3. This line will provide all the answers, in Final moles, to the problems.

Line 1 2 H2 + O2 → 2 H2O Line 2 Starting moles 2 2 0 Line 3 Reacting moles -2x -1x +2x Line 4 Final moles 2-2x 2-1x 0+2x

Figure 24 Sample SRF Table When the reaction is completed, either the hydrogen will be consumed or the oxygen will be consumed. That means that either 2 – 2x = 0 (hydrogen) or 2 - 1x = 0 (oxygen) If we solve for x in both equations the smallest x value will be the limiting reactant. The larger value will represent an amount in excess of what is possible with the given ingredients. So for this example, x = 1 (hydrogen) or x = 2 (oxygen). Since x = 1 is smallest, it is the value we choose. This step also identifies the substance that will run out first, the limiting reactant. In this case the hydrogen is the limiting reactant. Using the value of x = 1, we can determine the final amounts. The amount of product is 2x = 2(1) = 2 moles of H2O. The reactant in excess is O2 and it is excess by 2 – 1x = 2 – 1(1) = 1 mole of O2. This method of working the problem gives the same result as the visual inspection—hydrogen is the limiting reactant, and 2.00 mol H2O are produced. The advantage of learning this method is that it works even when the coefficients become difficult to use with the visual method. Example

Aluminum chloride, AlCl3, has many uses including in deodorants and antiperspirants. It is synthesized from aluminum and chlorine. What mass of AlCl3 can be produced if 100 g of each reactant are available? What is the limiting reactant? How many grams of the excess reactant remain? Step 1: Write the balanced equation.

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)

Step 2: Set up the SRF Table. Since only moles can go into the table, the grams of each reactant will first need to be converted to moles. See Figure 24.

22

22 41.1

gCl 71.0Cl mol 1Cl g 100

70.3Al g 27.0Al mol 1 Al g 100

Clmoles

Almoles

=×

=×

2 Al(s) + 3 Cl2(g) → 2 AlCl3(s) Starting moles 3.70 1.41 0 Reacting moles -2x -3x +2x Final moles 3.70 - 2x 1.41 –3x 2x

Step 3: Set the “Final moles” equal to zero and solve for x. Choose the smallest value of x.

3.70 – 2x = 0 or 1.41 – 3x = 0


x = 1.85 moles or x = 0.47 moles

Since x = 0.47 mol is smaller, Cl2 will be consumed first; it is the limiting reactant. Step 4: Determine the amount of product formed and the amount of Al remaining by

substituting x = 0.47 into the corresponding equations on the “Final moles” line. Amount of product (AlCl3): 2x = 2(0.47 mole) = 0.94 moles AlCl3

Amount of Al remaining: 3.70 – 2(0.47 mole) = 2.76 moles of Al

Step 5: Convert the moles back to grams.

leftAlgrams74.5mole1grams27.0Almoles2.76

formedAlClgrams126mole1grams134AlClmoles0.94 33

=×

=×


Skill Building Topic 10 UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS Practice Exercises 1. Acetylene burns in air to form carbon dioxide and water:

5 C2H2(g) + 2 O2(g) → 4 CO2(g) + 2 H2O(l) How many moles of CO2 are formed from 25.0 moles C2H2?

2. If insufficient oxygen is available, carbon monoxide can be a product of the combustion of butane: 9

C4H10(l) + 2 O2(g) → 8 CO(g) + 10 H2O(l). What mass of CO could be produced from 5.0 g butane?

3. 15.0 g NaNH2 is required for an experiment. Using the following reaction, what mass of sodium is required for reaction.

2 Na(s) + 2 NH3(g) → 2 NaNH2(s) + H2(g)?

4. Ethanol and acetic acid react to produce ethyl acetate according to the reaction C2H5OH +

CH3COOH → CH3COOC2H5. If the reaction is only 35% efficient at the conditions used, what mass of CH3COOH will be necessary to produce 100 g CH3COOC2H5? Assume that sufficient ethanol is available.

5. Heating CaCO3 yields CaO and CO2. Write the balanced equation. Calculate the mass of CaCO3

consumed when 4.65 g of CaO forms.


6. In the synthesis of sodium amide (NaNH2), what is the maximum mass of NaNH2 possible if 50.0 g of Na and 50.0 g NH3 were used?

2 Na(l) + 2 NH3(g) → 2 NaNH2(s) + H2(g)

7. The fuel methanol, CH3OH, can be made directly from carbon monoxide (CO) and hydrogen (H2).

a. Write a balanced equation for the reaction.

b. Calculate the maximum mass of methanol if one starts with 5.75 g CO and 10.0 g H2.

c. Which reactant is the limiting reactant?

d. How much of the excess reactant remains? 8. Aspirin (C9H8O4) is synthesized in the laboratory from salicylic acid (C7H6O3) and acetic anhydride

(C4H6O3): C7H6O3(s) + C4H6O3(l) → C9H8O4(s) + CH3COOH(l)

a. What is the theoretical yield of aspirin if you started with 15.0 g salicylic acid and 15.0 g acetic anhydride?

b. Which reactant is the limiting reactant?

c. What mass of the excess reactant remains?

DAY FIVE: YOU ARE THE CHEMIST

Putting it all Together You are the Chemist Introduction As an Undergraduate Senior in a Chemical Science, you have been hired by the SPC Corporation to work at their Green River Wyoming Plant as an Engineering Summer Intern. You will be working with the Plant Engineer on the Solvay Process. You have done an internet search on this process and have found the following information: Solvay Process for the Production of Sodium Carbonate

History

Two Belgian brothers, Ernest and Alfred Solvay, perfected a new procedure for producing soda ash (sodium carbonate) in 1861. They licensed their patent for this procedure in several countries. The American patents were held by the Solvay Process Company, incorporated in New York in 1881. The company's first plant was near Syracuse, New York, near a limestone quarry that provided the essential raw materials. Other plants were built in Hutchinson, Kansas; Detroit; and Milwaukee. The manufacture of soda ash proved extremely profitable, as it was sold for a wide variety of industrial uses. Solvay eventually expanded into the production of other alkali products, including caustic soda (sodium hydroxide), salts, calcium chloride and baking soda (sodium bicarbonate). A second plant was opened in Detroit, Michigan in 1897. The Solvay Process Company was absorbed in 1921 by the Allied Chemical & Dye Corporation. Process Summary

Sodium carbonate, Na2CO3, soda ash, has a number of uses but its most common use is in the production of glass. Since the 1860's, sodium carbonate has been produced using the Solvay Process. The Solvay Process is a continuous process using limestone (CaCO3) to produce carbon dioxide (CO2) which reacts with ammonia (NH3) dissolved in brine (concentrated NaCl(aq)) to produce sodium carbonate (Na2CO3). The steps in the Solvay Process are:

1. Brine Purification (Sodium Chloride Purification) 2. Sodium Hydrogen Carbonate Formation 3. Sodium Carbonate Formation 4. Ammonia Recovery

Sodium carbonate, Na2CO3, dissolves in water to form an alkaline solution. Used as a base, sodium carbonate is cheaper and safer than sodium hydroxide.

http://www.ausetute.com.au/acidbase.html


Uses of Sodium Carbonate

Use Process Notes

Glass Making A mixture of Na2CO3, CaCO3 and SiO2 (silicon dioxide sand) is used for window or bottle glass.

Water Softening Agent

CO32- from dissolved Na2CO3 can precipitate

Mg2+ and Ca2+ ions from hard water as the insoluble carbonates, preventing them from forming a precipitate with soap resulting in scum. For this reason, sodium carbonate is also known as washing soda.

Paper Making Na2CO3 is used to produce the NaHSO3 necessary for the sulfite method of separating lignin from cellulose.

Baking Soda Production Baking soda (or sodium hydrogen carbonate or sodium bicarbonate), NaHCO3, is used in food preparation and in fire extinguishers.

Sodium Hydroxide Production for Soaps and Detergents

Na2CO3 is reacted with a Ca(OH)2, slaked lime, suspension.

Wool Processing Na2CO3 removes grease from wool and neutralises acidic solutions.

Power Generation Na2CO3 is used to remove SO2(g) from flue gases in power stations.

http://www.ausetute.com.au/soaps.html

http://www.ausetute.com.au/titrcalc.html


Solvay Process

The Solvay Process for the production of sodium carbonate is summarised in the flowchart below:

brine NaCl(aq)

----->ammoniated brine<-----ammonia

NH3

| |

/\ |

limestone CaCO3

| | | |

NaCl H2O NH3

| | | |

NH3

| | \/

| | | \/

| | | |

lime kiln CO2

----->carbonating tower

| | |

H2O

| CaO

| \/

| |

| \/

| \/

filter

| NH4Cl

---------> lime

slaker Ca(OH)2 -------------|--------------->

ammonia recovery

| | \/

| \/

product NaHCO3

by-product CaCl2

| 300oC

| \/

product Na2CO3


Process Steps

Brine Purification

Brine (Sodium Chloride solution) is concentrated by evaporation to at least 30% Impurities such as calcium, magnesium and iron are removed by precipitation as insoluble salts:

Ca2+(aq) + CO3

2-(aq) -----> CaCO3(s)

Mg2+(aq) + 2OH-

(aq) -----> Mg(OH)2(s) Fe3+

(aq) + 3OH-(aq) -----> Fe(OH)3(s)

Brine solution is then filtered and passed through an ammonia tower to dissolve ammonia. This process is exothermic, releases energy, so the ammonia tower is cooled.

Sodium Hydrogen Carbonate Formation

Carbon dioixide is produced by the thermal decomposition of limestone, CaCO3(s), in the lime kiln:

CaCO3(s) -----> CO2(g) + CaO(s)

Carbon dioxide is bubbled through the ammoniated brine solution in the carbonating tower. The carbon dioxide dissolves to form a weak acid:

CO2(g) + H2O(l) HCO3-(aq) + H+

(aq)

The ammonia in the brine reacts with H+ to form ammonium ions:

NH3(aq) + H+(aq) NH4

+(aq)

The HCO3- then reacts with the Na+ to form a suspension of sodium hydrogen

carbonate:

HCO3-(aq) + Na+

(aq) NaHCO3(s)

NaHCO3 precipitates because of the large excess of Na+ present in the brine which forces the equilibrium position to shift to the right by Le Chatelier's Principle (NaHCO3 is quite soluble in water). The overall molecular equation for the formation of sodium hydrogen carbonate in the carbonating tower is:

NH3(aq) + CO2(g) + NaCl(aq) + H2O(l) -----> NaHCO3(s) + NH4Cl(aq)

The net ionic equation for the formation of sodium hydrogen carbonate:

NH3(aq) + CO2(g) + Na+(aq) + H2O(l) -----> NaHCO3(s) + NH4

+(aq)

where Cl- is a spectator ion (it does not participate in the reaction).

http://www.ausetute.com.au/solrules.html

http://www.ausetute.com.au/enthchan.html

http://www.ausetute.com.au/ppteeqtn.html

http://www.ausetute.com.au/lechatsp.html

http://www.ausetute.com.au/lechatsp.html


Sodium Carbonate Formation

Suspended sodium hydrogen carbonate is removed from the carbonating tower and heated at 300oC to produce sodium carbonate:

2NaHCO3(s) -----> Na2CO3(s) + CO2(g) + H2O(g)

The carbon dioxide produced is recycled back into the carbonating tower.

Ammonia Recovery

CaO is formed as a by-product of the thermal decomposition of limestone in the lime kiln. This CaO enters a lime slaker to react with water to form calcium hydroxide:

CaO(s) + H2O(l) -----> Ca(OH)2(aq)

The calcium hydroxide produced here is reacted with the ammonium chloride separated out of the carbonating tower by filtration:

Ca(OH)2(aq) + 2NH4Cl(aq) -----> CaCl2(aq) + 2H2O(l) + 2NH3(g)

The ammonia is recycled back into the process to form ammoniated brine. Calcium chloride is formed as a by-product of the Solvay Process.

Environmental Issues

1. Solid Wastes

Calcium chloride, CaCl2, is a by-product of the Solvay Process. There are a limited number of uses for CaCl2 as a drying agent in industry, de-icing roads, an additive in soil treatment, and an additive in concrete. The rest must be disposed of by evaporating to dryness and disposing of the solid. CaCl2 can not be pumped into the Green River because it will raise the concentration of chloride ion to unacceptable levels. Other solid wastes include unreacted calcium carbonate, sand and clay from the kiln. These are used to make bricks, landfill, or road base.

2. Air Pollution

Some ammonia is lost to the atmosphere during the Solvay Process. Ammonia is a toxic atmospheric pollutant. Ammonia losses are minimized by recycling the gas to reduce plant operation costs. Carbon dioxide gas is also recycled.

3. Thermal Pollution

Some of the processes involved in the Solvay Process are exothermic, they release heat. The Green River Plant has to cool the water first before releasing in order to prevent disruption to aquatic organisms.


Based on the manufacturing information in this packet and in the skills that you learned in the Skill Building assignments answer the following questions:

1. If SPC’s Green River Plant produces 325,000,000 Kg per year of sodium carbonate. How many moles of calcium carbonate are needed per year to produce this? Take the overall equation as,

CaCO3(s) + 2NaCl (aq) Na2CO3(aq) + CaCl2(aq)

Would this problem have been easier for you to do the calculation if the scientific notation value of 3.25 x 108 Kg had been used?

2. Evaporative basins at Dry Creek produce an average of 650,000,000 Kg per year of salt. This is purified, then dissolved to form a saturated brine solution that is pumped to the Solvay plant.

Ammonia is dissolved in the brine solution and then the ammoniated brine is reacted with carbon dioxide.

A. Write an equation for this reaction. B. If 50% of the original salt is sodium chloride, what mass of ammonia

will be needed to react with it? C. Since this is a continuous process and almost all of the ammonia used is

continuously recycled and reused many times on a daily basis, is the amount calculated in B realistic?

3. How much carbon dioxide must be made per year by heating calcium carbonate? (Hint: one mole of carbon dioxide is produced for each mole of calcium carbonate used.)

4. Can all of the waste carbon dioxide from the last step in the process be reused in the carbonating tower? Since this is a continuous process, is this amount of carbon dioxide the same amount as calculated in problem 4 or is it very much less? Explain your answer in terms of the Sodium Carbonate formation reaction and the nature of recycling materials in a continuous process.

2NaHCO3(s) -----> Na2CO3(s) + CO2(g) + H2O(g)

Periodic Table of the Elements Chemistry Reference Sheet California Standards Test

Sodium22.99

Na11 Atomic number

Element symbol

Average atomic mass* Element name

Hydrogen1.01

H1

Lithium6.94

Li3

Sodium22.99

Na11

Potassium39.10

19

K Nickel58.69

Ni28

Rubidium85.47

Rb37

Rutherfordium

(261)

Rf104

Molybdenum

95.94

Mo42

Germanium72.61

Ge32

11A

22A

1

2

3

4

77B

111B

122B

133A

166AKey

8

5

6

7

98B

10

144A

155A

177A

188A

33B

44B

55B

66B

Copper63.55

Cu29

Cobalt58.93

Co27

Helium4.00

He2

Boron10.81

B5

Carbon12.01

C6

Nitrogen14.01

N7

Oxygen16.00

O8

Fluorine19.00

F9

Neon20.18

Ne10

Aluminum26.98

Al13

Silicon28.09

Si14

Phosphorus30.97

P15

Sulfur32.07

S16

Chlorine35.45

Cl17

Argon39.95

Ar18

Calcium40.08

Ca20

Scandium44.96

Sc21

Titanium47.87

Ti22

Chromium52.00

Cr24

Iron55.85

Fe26

Zinc65.39

Zn30

Gallium69.72

Ga31

Arsenic74.92

As33

Selenium78.96

Se34

Bromine79.90

Br35

Krypton83.80

Kr36

Strontium87.62

Sr38

Yttrium88.91

Y39

Zirconium91.22

Zr40

Niobium92.91

Nb41

Technetium(98)

Tc43

Ruthenium101.07

Ru44

Rhodium102.91

Rh45

Palladium106.42

46

Silver107.87

Ag47

Cadmium112.41

Cd48

Indium114.82

In49

Tin118.71

Sn50

Antimony121.76

Sb51

Tellurium127.60

Te52

Iodine126.90

I53

Xenon131.29

Xe54

Cesium132.91

Cs55

Barium137.33

Ba56

Lanthanum138.91

La57

Hafnium178.49

Hf72

Tantalum180.95

Ta73

Tungsten183.84

W74

Rhenium186.21

Re75

Osmium190.23

Os76

Iridium192.22

Ir77

Platinum195.08

Pt78

Gold196.97

Au79

Mercury200.59

Hg80

Thallium204.38

Tl81

Lead207.2

Pb82

Bismuth208.98

Bi83

Polonium(209)

Po84

Astatine(210)

At85

Pd

Radon(222)

Rn86

Francium(223)

Fr87

Radium(226)

Ra88

Actinium(227)

Ac89

Dubnium(262)

Db105

Seaborgium(266)

Sg106

Bohrium(264)

Bh107

Hassium(269)

Hs108

Meitnerium(268)

Mt109

Magnesium24.31

Mg12

Beryllium9.01

Be4

Vanadium50.94

V23

Manganese54.94

Mn25

* If this number is in parentheses, then it refers to the atomic mass of the most stable isotope.

Praseodymium

140.91

Pr59

Mendelevium

(258)

Md101

Cerium140.12

Ce58

Neodymium144.24

Nd60

Promethium(145)

Pm61

Samarium150.36

Sm62

Europium151.96

Eu63

Gadolinium157.25

Gd64

Terbium158.93

Tb65

Dysprosium162.50

Dy66

Holmium164.93

Ho67

Erbium167.26

Er68

Thulium168.93

Tm69

Ytterbium173.04

Yb70

Lutetium174.97

Lu71

Thorium232.04

Th90

Protactinium231.04

Pa91

Uranium238.03

U92

Neptunium(237)

Np93

Plutonium(244)

Pu94

Americium(243)

Am95

Curium(247)

Cm96

Berkelium(247)

Bk97

Californium(251)

Cf98

Einsteinium(252)

Es99

Fermium(257)

Fm100

Nobelium(259)

No102

Lawrencium(262)

Lr103

Copyright © 2008 California Department of Education

Formulas Ideal Gas Law: PV = nRT Calorimetric Formulas –

P1V1 P2V2Combined Gas Law: = No Phase Change: Q = m(ΔT)CpT1 T2

Pressure Formula: P= F Latent Heat of Fusion: Q = mΔHfusA

Mass-Energy Formula: E = mc 2 Latent Heat of Vaporization: Q = mΔHvap

Constants LVolume of Ideal Gas at STP: 22.4

mol

Speed of Light in a Vacuum: c = 3.00 × 108 ms

Specific Heat of Water: Cp(H 2O) = 1.00 cal J (g C)

= 4.18 (g C)

cal JLatent Heat of Fusion of Water: ΔHfus(H 2O) = 80 g = 334 g

cal JLatent Heat of Vaporization of Water: ΔHvap(H 2O) = 540 g = 2260 g

Unit Conversions Calorie-Joule Conversion: 1 cal = 4.184 J

Absolute Temperature Conversion: K = C + 273

lbs.Pressure Conversions: 1 atm = 760 mm Hg = 760 Torr =101.325 kPa = 14.7 in.2 = 29.92 in. Hg

Formulas, Constants, and Unit Conversions Chemistry Reference Sheet California Standards Test

Copyright © 2008 California Department of Education

Day Four – Honors Chemistry Addendum

In a laboratory after the proper mass measurements had been made, a sample of pure nickel was heated in crucible, the nickel reacted with oxygen. Once the reaction was complete the product formed was nickel oxide, the mass was taken and the following data was collected

Mass of crucible = 30.02

Mass of Nickel and Crucible = 31.07

Mass of Nickel Oxide and crucible = 31.36

Determine the following information based on the data given

Mass of nickel = _____________________

Mass of nickel oxide = ____________________

Mass of oxygen = _______________________

Based on these calculations what is the empirical formula for the nickel oxide? ________________

1. Write the chemical formula for the reactants in the above reaction.

2. Write chemical formula for the product in the above reaction.

3. Describe what happened in the above reaction.

Day Five ‐ Honors Chemistry Addendum

1. Look up if you do not remember the equation for respiration and fill in the blanks

_____________ + _______________ _______________ + ______________

2. Make sure the equation is balanced if you have not done so already.

3. The average person consumes 0.84kg of oxygen per day, how many liters of Carbon dioxide will be produced by the average person in one day?

4. If there is 2200grams of glucose and 840g of Oxygen available what is the limiting reagent?

5. Using the limiting reagent from problem 4 calculate the amount of Carbon dioxide produced and the amount of water produced.

6. Using problems 4 and 5 as evidence, explain why the intake of food is necessary.

laboratory chemistry emergency plan days 6 -...

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