Post on 12-Mar-2018
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 %
=−=
=×=
DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS
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
=
=
DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS
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.
DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS
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.
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4. Find the percent of each element in the following compounds. a. MgSO4
b. MgSO4 • 7 H2O
c. (CH3)2SO
d. C5H5N
DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS
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?
DAY FOUR: A QUANTITATIVE UNDERSTANDING OF FORMULAS
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
DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS
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.
DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS
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
DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS
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
=×
=×
DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS
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.
DAY FIVE: UNDERSTANDING MASS RELATIONSHIPS IN CHEMICAL REACTIONS
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.
DAY FIVE: YOU ARE THE CHEMIST
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.
DAY FIVE: YOU ARE THE CHEMIST
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
DAY FIVE: YOU ARE THE CHEMIST
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).
DAY FIVE: YOU ARE THE CHEMIST
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.
DAY FIVE: YOU ARE THE CHEMIST
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.