Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 1. A chemical equation. (C4.4)


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1 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question What is a symbolic representation of a chemical reaction? What 3 things (values) is a mole of a chemical equal to? What is the average mass (in atomic mass units) of the naturally occurring isotopes of an element? What is the number named for Amadeo Avogadro and what is that number equal to? What is the symbol for atomic mass unit? The typical formula for oxygen is? The typical formula for fluorine is? The typical formula for hydrogen is? The typical formula for iron is? The typical formula for argon is? 7. What is an equality statement? Answer A chemical equation. (C4.4) There are 3 answers that you should know so far from the unit handout and your mole day project: (C4.4) a) 1 mole of any elemental particle (atoms, molecules, formula units, electrons, neutrons, protons, monoatomic and polyatomic ions, and so on) is equal to 6.02 x of that elemental particle. i.e x of an object = 1 mole of those same objects. b) 1 mole of any gas is equal to 22.4 liters of that gas but only so long as the gas is at standard temperature and pressure (STP) c) 1 mole of any chemical is equal to the sum of the atomic masses of all the atoms in the formula unit of that chemical with those masses written in units of grams Atomic mass. (C4.4) and of anything equals a mole of those things. (C4.4) u is the preferred symbol for atomic mass unit but amu is an older symbol. (C4.4) O 2. F 2. H 2. Fe. Ar. A mathematic statement showing that 2 values with different measurement units are equal to each other. Example: 22.4 L of gas = 1 mole of gas. (C4.4) More information page 1. Every student was required to complete a mole project in which this question was answered. This can also be found on the Mole Map on page 23 of the unit handout. page 3. page 3. page 3. Previous units beginning on page 7 but mentioned throughout.
2 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 2 Question 8. What is formula mass? 9. What is molecular mass? Where do we get the molar ratio (be very specific)? What is molar mass and how do we in this class calculate it? What is the maximum amount of product that can be produced from a given amount of limiting reactant? What do we in this class use an equality statement to construct? Answer Formula mass is the term primarily used for ionic substances. It is the sum of the atomic masses of all of the elements contained in one formula unit of a compound; typically an ionic compound. (C4.4) Molecular mass is the term used for molecular compounds. It is the sum of the atomic masses of all of the elements in the molecular formula of the substance. (C4.4) The numerical values in molar ratios come from the coefficients in a balanced chemical equation. (C4.4) Molar mass is the sum of the atomic masses of the elements in a formula unit, molecular formula, or other chemical formula written with units of grams. This calculation is equal to one mole of that formula unit. There is a required format for calculating molar mass which shows the elements in the formula, the atomic masses of those elements, a multiplicands which is numerically equal to the number of atoms of that element in the formula, the product of those atomic masses and multiplicand, the sum of the products, units of grams and the species for the formula and a equality statement which shows the mass equal to a mole of that same species. (C4.4) Theoretical yield. (C4.4) An equality statement is used to construct conversion units. (C4.4) More information page 6. page 6. beginning on page 13 but mentioned throughout. Throughout the unit handout; but there is a clear example on page 10. page 19. Throughout the unit handout; but there is a clear example on page 7.
3 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question What is a limiting reactant and how do we determine which reactant is the limiting reactant? 15. What is percent yield? 16. The clean surface of pure aluminum quickly oxidizes to form a coating of aluminum oxide (Al 2 O 3 ) that actually protects the aluminum under the surface from further corrosion. What is the percent composition of aluminum oxide? Answer A limiting reactant is the reactant that runs out 1 st in the chemical reaction, leaving some amount of the excess reactant. In this unit, we determine the limiting reactant by performing a masstomass calculation from the given amount of one of the reactants to a calculated amount of the 2 nd reactant. If the calculated mass of the 2 nd reactant is greater than the given mass of that reactant, than that same reactant is the limiting reactant (the given mass is the mass in the problem or the mass that you measure in the lab). If not, the 2 nd reactant is the excess reactant and the 1 st reactant is the limiting reactant. (C4.4) Percent yield is the actual yield divided by the theoretical yield multiplied by 100. (C 4.4) % Al and % O. (C4.5) More information page 16 and 17. page 20. page 10. Also, see the complete explanation for this calculation below.
4 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page Question You perform the following lab procedure: 2.00g of calcium chloride and 1.00g of aluminum sulfate are dissolved in separate containers with 100mL of water which are then mixed together. A white mixture forms and the mixture is allowed to sit, undisturbed, for 10 minutes. After sitting it is observed that the white material has settled to the bottom of the beaker and a clear liquid remains in the beaker. The clear liquid is decanted and the white material is filtered, rinsed, dried, and weighed. The amount of product recovered weighs 0.45g. What is the solid, white product, what is the percent yield, and what is the amount of excess reactant that remains in solution? So long as there is plenty of oxygen, burning of hydrocarbon fuels produces carbon dioxide. When the supply of oxygen is limited however, burning of hydrocarbon fuels produces poisonous carbon monoxide. Propose an explanation for this phenomenon. Answer Solid, white product is CaSO 4. % yield is 38%. Amount of excess reactant remaining is 1.03 g CaCl 2. (C4.5) More information pages 17 through 22. Also, see the complete explanation for this calculation below. Since the reaction with a hydrocarbon fuel and oxygen can produce water and EITHER carbon dioxide and carbon monoxide there must be a model that would explain why one or the other might be produced. One possible model (and it is by no means the ONLY reasonable model) is this: When the oxygen is limited, the carbon will combine with oxygen in a form that uses less oxygen. CO (carbon monoxide) has less oxygen per molecule or per mole than CO 2 (carbon dioxide).
5 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 5 Problem #16; complete explanation: What is the percent composition of aluminum oxide (Al 2 O 3 )? Percent composition simply means the percent mass of each element in a compound. Since we were using carbon dioxide as an example earlier, let s continue with that for now. The atomic mass of aluminum (elemental symbol Al) is u and the atomic mass of oxygen (elemental symbol O) is u (see the periodic table in your test references; these numbers are in the bottom rectangle for these elements). The sum of the atomic masses of both aluminum atoms in aluminum oxide is u. The sum of the atomic masses of the 3 oxygen atoms in aluminum oxide is u. The total atomic mass of all the atoms in the formula (the sum of the atomic masses) for aluminum oxide is u which must be rounded to u. To find the percent composition for aluminum you simply take the total mass of the 2 aluminum atoms and divide by the entire mass of aluminum oxide formula unit then multiply by 100. To find the percent composition for oxygen you simply take the total mass of the 3 oxygen atoms and divide by the entire mass of aluminum oxide formula unit then multiply by 100. The percent composition for aluminum oxide comes in 2 parts: 1) the percent mass of aluminum and 2) the percent mass of oxygen. The percent composition for any compound comes in as many parts as there are elements in the formula. Example: Percent composition of aluminum oxide Example: The required format for calculating percent composition begins with the same format required for calculating molar mass in this class (meaning, if you don t do it I count off points of tests, homework, and class work). So for carbon dioxide you begin with The required format for calculating molar mass and the beginning of percent composition in this class. Al: = Mass of all aluminum atoms O: = Mass of all oxygen atoms g Al 2 O g Al 2 O 3 = 1 mol Al 2 O 3 Mass of the aluminum oxide formula unit Equality statement showing that the calculated mass of aluminum oxide is equal to a mole of aluminum oxide. From the calculations above, you have the mass of all the aluminum atoms and the mass of all the oxygen atoms and the mass of the entire compound. % mass of aluminum in Al O = 2 3 % mass of oxygen in Al O = u u u u % % Al in Al O % % O in Al2O 3. Check: The sum of the percent mass of aluminum and the percent mass of oxygen is equal to or is very close to 100%: % % % % 2 3
6 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 6 Problem #17; complete explanation: What is the solid, white product, what is the percent yield, and what is the amount of excess reactant that remains in solution? 2.00g of calcium chloride and 1.00g of aluminum sulfate dissolved in separate containers of water, a white precipitate forms and is collected and weighed. The amount of product recovered weighs 0.45g. Step 1: Make sure the chemical equation is balanced. If you have an unbalanced chemical equation presented to you, then you need to balance the equation first. If you are only given the reactants you will have to predict the products and balance the product formulas (you cannot properly balance a chemical equation unless the formulas are balanced). o You may only be given the word formulas for the reactants, in which case you must be able to write the symbolic formula from the word formula and balance those formulas as well. o In order to predict products you must know how to classify chemical equations. Calcium has a 2+ charge and chloride has a 1 charge (see your periodic table of oxidation numbers in your test references). So the balanced formula for this compound is CaCl 2. Aluminum has a 3+ charge and sulfate has a 2 charge (see your common ions chart in your test references to find the formula and charge on the sulfate polyatomic ion). So the balanced formula for this compound is Al 2 (SO 4 ) 3. Review unit 4 if you don t recall how balance and write formulas. So the beginning of our equation is CaCl 2 + Al 2 (SO 4 ) 3. Both reactants are soluble (see solubility rules in test references), so CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq). Both formulas are in 2 parts, so the likelihood is that this will be a double replacement reaction. The predicted products in this unbalanced equation will be CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq) AlCl 3 (aq) + CaSO 4 (s). Using the solubility rules you should be able to determine that aluminum chloride is soluble and since this reaction occurred in a water (aqueous) environment it will be identified as aqueous (aq). Calcium sulfate is not soluble and will, therefore, be a solid(s) and must be identified as such. This answers the 1 st part of the question, What is the solid, white product? Answer: calcium sulfate or CaSO 4. To balance the equation (or just to show that the equation is already balanced) you MUST have an atom inventory. The result is 3CaCl 2 (aq) + Al 2 (SO 4 ) 3 (aq) 2AlCl 3 (aq) + 3CaSO 4 (s). Ca Cl Al S O Review unit 5 if you don t recall how predict products or balance equations.
7 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 7 Step 2: Find the limiting reactant (also called the limiting reagent or limiting factor). In this unit, this determination will always be a masstomass calculation. This masstomass calculation will start with one reactant and end with the other reactant. o The rule is that if the calculated amount (called a theoretical amount) of the reactant at the end of the masstomass calculation is greater than the amount given in the problem, then the species at the end of the calculation is the limiting reactant. o If the calculated amount (called a theoretical amount) of the reactant at the end of the masstomass calculation is smaller than the amount given in the problem, then the species at the end of the calculation is the excess reactant. o One reactant is virtually always a limiting reactant and the other reactant is the excess reactant. To find a limiting reactant you must know how to calculate the molar mass (or molar weight) of the reactants and products AND you must know how to determine a molar ratio. o Determine molar mass by writing out the formula, such as H 2 O for water, and look up the weight of each atom on the periodic table. For example, multiply the hydrogen atom weight by two and add it to the weight of oxygen. Use only the molar mass calculation format required and approved for this class. o The molar ratio comes from the coefficients and formulas in the balanced chemical equation. Perform a masstomass calculation from one reactant to another. To do this we need to calculate the molar mass of calcium chloride and aluminum sulfate: Ca: = Cl: = g CaCl 2 = 1 mol acl 2 Now that we have this equality statement ( g CaCl 2 = 1 mol CaCl 2 ) we can enter a molar conversion unit for calcium chloride to the calculation. Al: = S: = O: = g Al 2 (SO 4 ) 3 = 1 mol Al 2 (SO 4 ) 3 Now, using these equality statements and the molar ratio from the balanced equation we can construct the following masstomass calculation from the mass for calcium chloride that is given in the problem to the mass of aluminum sulfate needed to completely react with that amount of calcium chloride g CaCl2 1 mol CaCl 2 1 mol Al SO g Al SO g CaCl2 3 mol CaCl 2 1 mol Al SO g Al SO 2.06 g Al SO. This masstomass calculation demonstrates that you need 2.06 grams of aluminum sulfate to completely react with all of the 2.00 grams of calcium chloride that you were given. The problem is that you don t have need 2.06 grams of aluminum sulfate. You were only given 1.00 gram of
8 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 8 aluminum sulfate. Therefore, aluminum sulfate must be the limiting reactant. Every masstomass calculation that follows this one will start with the given amount (1.00 gram) of aluminum sulfate. Step 3: Calculate the theoretical yield or how much will be synthesized under perfect conditions. In this unit, this determination will always be a masstomass calculation. To do this we must perform a masstomass calculation from the given amount of limiting reactant to the product that we are seeking. The product we are seeking is clearly identified as the solid, white precipitate (read the original question), and that is calcium sulfate (CaSO 4 ). Rule for determining the limiting reactant: 1. Complete a masstomass calculation from one reactant to the other. 2. If the amount of reactant that you calculated is greater than the amount given to you in the problem the ending species is the limiting reactant. 3. If the amount calculated is less than the amount given to you in the problem the ending species is the excess reactant. To perform this masstomass calculation, we will need a molar mass calculation for calcium sulfate: Ca: = S: = O: = g CaSO 4 = 1 mol CaSO g Al SO 1 mol 1 Al SO g Al SO 3 mol CaSO4 1 mol Al SO g CaSO 4 1 mol CaSO g CaSO 1.19 g CaSO. 4 4 Step 4: Know the actual yield, or the amount of product truly synthesized in the original experiment. This involves no additional calculations. It is simply the amount measured at the end of the experiment or given to you in a word problem. The problem states that, The amount of product recovered weighs 0.45 g. This, then, is the actual yield. Step 5: Calculate percentage yield (mass of actual yield divided by mass of theoretical yield multiplied by 100 percent). Percent yield is the actual yield divided by the theoretical yield multiplied by 100. Actual Yield Theoretical Yield 100 = Percentage Yield Therefore, the percentage yield would be: 0.45 g CaSO g CaSO4 100 = % 37% yield in calcium sulfate. Step 6: Calculate the amount of excess reactant remaining after the reaction is complete.
9 Honors Chemistry: Unit 6 Test Stoichiometry PRACTICE TEST ANSWER KEY Page 9 Before we can calculate the amount of excess reactant remaining, we must first determine the amount of excess reactant used. This also involves a masstomass calculation. For this calculation we also start with the limiting reactant, but we calculate to the other reactant the excess reactant. This allows us to determine just how much of the excess reactant is actually used up in the reaction: 1.00 g Al SO 1 mol 1 Al SO g Al SO 3 mol CaCl2 1 mol Al SO g CaCl 2 1 mol CaCl g CaCl g CaCl. 2 2 To find the amount of excess reactant remaining, we simply take this amount (0.973 g CaCl 2 ) and subtract it from the amount of calcium chloride given in the problem: 2.00 g CaCl 2 given in the problem g CaCl 2 used up in the reaction g CaCl g CaCl 2 excess reactant remaining after the reaction is finished.
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