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Chemistry

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Jan 9, 2024

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Name(s): __________________ CHM 1020 W5 Week 5 Assignment Objective In today’s lab, we will demonstrate a few different types of chemical reactions and be able to observe whether a chemical reaction has taken place. We will also learn how to use chemical equations to describe chemical reactions. Background/Theory The Periodic Table shows over 100 elements. The chemical literature describes millions of compounds that are known—some isolated from natural sources, some synthesized by laboratory workers. The combination of chemicals, in the natural environment or the laboratory setting, involves chemical reactions. The change in the way that matter is composed is a chemical reaction , a process wherein reactants (or starting materials) are converted into products. The new products often have properties and characteristics that are entirely different from those of the starting materials. Four ways in which chemical reactions may be classified are combination, decomposition, single replacement (substitution), and double replacement (metathesis). Two elements reacting to form a compound is a combination reaction . This process may be described by the general formula: A + B AB The rusting of iron or the combination of iron and sulfur are good examples. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) (rust) Fe(s) + S(s) FeS(s) Two compounds reacting together as in the example below also is a combination reaction. CaO(s) + CO 2 (g) CaCO 3 (s) A compound which breaks down into elements or simpler components typifies the decomposition reaction . This reaction has the general formula: AB A + B Some examples of this type of reaction are the electrolysis of water into hydrogen and oxygen: 2H 2 O(l) 2H 2 (g) + O 2 (g) and the decomposition of potassium iodate into potassium iodide and oxygen: 2KIO 3 (s) 2KI(s) + 3O 2 (g) The replacement of one component in a compound by another describes the single replacement (or substitution ) reaction. This reaction has the general formula: AB + C CB + A Page 1 of 9
Name(s): __________________ CHM 1020 W5 Processes which involve oxidation (the loss of electrons or the gain of relative positive charge) and reduction (the gain of electrons or the loss of relative positive charge) are typical of these reactions. Use of Table 1, the activity series of common metals, enables chemists to predict which oxidation-reduction reactions are possible. A more active metal, one higher in the table, is able to displace a less active metal, one listed lower in the table, from its aqueous salt. Thus aluminum metal displaces copper metal from an aqueous solution of copper(II) chloride; but copper metal will not displace aluminum from an aqueous solution of aluminum(III) chloride. 2Al(s) + 3CuCl 2 (aq) 3Cu(s) + 2AlCl 3 (aq) Cu(s) + AlCl 3 (aq) No Reaction ( Note that Al is oxidized to Al 3+ and Cu 2+ is reduced to Cu. ) Hydrogen may be displaced from water by a very active metal. Alkali metals are particularly reactive with water, and the reaction of sodium with water often is exothermic enough to ignite the hydrogen gas released. 2Na(s) + 2HOH(l) 2NaOH(aq) + H 2 (g) + heat (Note that Na is oxidized to Na + and H + is reduced to H 2 .) Active metals, those above hydrogen in the series, are capable of displacing hydrogen from aqueous mineral acids such as HCl or H 2 SO 4 ; however, metals below hydrogen will not replace hydrogen. Thus zinc reacts with aqueous solutions of HCl and H 2 SO 4 to release hydrogen gas, but copper will not. Zn(s) + 2HCl(aq) ZnCl 2 (aq) + H 2 (g) Cu(s) + H 2 SO 4 (aq) No reaction Two compounds reacting with each other to form two different compounds describes double replacement reactions. This process has the general formula: AB + CD AD + CB There are two replacements in the sense that A replaces C in CD and C replaces A in AB. This type of reaction generally involves ions which form in solution either from the dissociation of ionic compounds or the ionization of molecular compounds. The reaction of an aqueous solution of silver nitrate with an aqueous solution of sodium chloride is a good example. The products are sodium nitrate and silver chloride. We know a reaction has taken place since the insoluble precipitate silver chloride forms and separates from solution. AgNO 3 (aq) + NaCl(aq) NaNO 3 (aq) + AgCl(s) (White precipitate) In general, a double replacement results if one combination of ions leads to a precipitate, a gas or an un- ionized or very slightly ionized species such as water. In all of these reaction classes, it is very often possible to use your physical senses to observe whether a chemical reaction has occurred. The qualitative criteria may involve the formation of a gaseous product, the formation of a precipitate, a change in color, or a transfer of energy. In the next section, you will practice identifying which of the four classes or classification of chemical reactions mentioned above does each chemical reaction represent. Note, in addition to combination, decomposition, single replacement and double replacement reactions, we also have combustion Page 2 of 9
Name(s): __________________ CHM 1020 W5 reactions, which you will explore in later assignments. For now, each reaction listed will fall into one of the four categories described here. Practice determining reaction classification For each of the reactions below, classify as a combination, decomposition, single replacement, or double replacement. 1. Ca(s) + Cl 2 (g) CaCl 2 (s) combination 2. 2Cu(s) + O 2 (g) 2CuO(s) combination 3. Ca(NO 3 ) 2 (aq) + H 2 SO 4 (aq) 2HNO 3 (aq) + CaSO 4 (s) Double replacement 4. NH 3 (aq) + HCl(aq) NH 4 Cl(aq) combination 5. Hg(NO 3 ) 2 (aq) + 2NaI(aq) HgI 2 (s) + 2NaNO 3 (aq) double replacement 6. AgNO 3 (aq) + NaCl(aq) AgCl(s) + NaNO 3 (aq) Double replacement 7. Zn(s) + H 2 SO 4 (aq) ZnSO 4 (aq) + H 2 (g) double replacement Page 3 of 9
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Name(s): __________________ CHM 1020 W5 8. H 2 CO 3 (aq) CO 2 (g) + H 2 O(l) single replacement 9. 2H 2 O(l) 2H 2 (g) + 2O 2 (g) decomposition 10. 2Li(s) + 2H 2 O(l) 2LiOH(aq) + H 2 (g) double replacement Procedure for Experiments Combination/Synthesis Reactions Watch the following Video : https://www.youtube.com/watch?v=w2ydd9rJHws 1. For the reaction of magnesium and atmospheric (diatomic) oxygen write out the balanced chemical equation: 2 Mg(s) + O 2 (g) 2 MgO(s) 2. Next describe what magnesium metal looks like, and what the product identified above looks like. For example, was the magnesium reactant shiny or dull, was it malleable or brittle? Magnesium reactant is silvery-white in colour and very light in weight. It's shiny malleable, and ductile. While product Magnesium oxide appears as white solid, hygroscopic in nature Decomposition Reaction Watch the following video: https://www.youtube.com/watch?v=_Y1alDuXm6A 1. Write out the balanced chemical equation for the thermal decomposition of Mercury oxide: Page 4 of 9
Name(s): __________________ CHM 1020 W5 2 HgO (s) --------> 2 Hg (s) + O 2 2. What did the Mercury (II) oxide reactant look like, and what did the products of the decomposition reaction look like? Reactant: The reactant was Mercury (II) oxide, and it looked like a red liquid. It was a solid at room temperature, but once heated up it became a gas. Products: Mercury is silvery-white in color, has a metallic luster like that of silver or gold, and is a very heavy metal. It is also highly toxic if ingested or inhaled. The other product of this reaction was oxygen gas; this gas is colorless and odorless at room temperature but becomes a bright blue when exposed to light or flame! 3. How do we know the gas released was oxygen? The sample gas can be isolated and then sparked to induce ignition. Since O 2 can undergo combustion, ignition should be observed if it is present. Single Displacement Reaction Watch the following video: https://www.youtube.com/watch?v=k8UtR7akNec 1. Write out the balanced chemical equation for single replacement reaction of silver nitrate and copper: 2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq) 2. Describe your observation, and clearly identify the appearances of reactants and products. There is a color change. The colorless solution of silver nitrate changes to a blue solution. This due to the reduction of copper to copper (II) ions which are blue in color. Copper displace silver from silver nitrate to form copper nitrate and silver solid. The presence of copper (II) ions in copper nitrate solution Page 5 of 9
Name(s): __________________ CHM 1020 W5 solution is responsible for the blue color of the solution. In addition, there is formation of a white solid (which is silver). Silver ions in silver nitrate are oxidized to silver metal which is a white solid. Double Displacement Reaction Watch the following video: https://www.youtube.com/watch?v=X2mB-q2NQXY 1. Write out the balanced chemical equation for the double replacement reaction of lead nitrate and potassium iodide: Pb(NO 3 ) 2 (aq) + 2 KI (aq) 2 KNO 3 (aq) + PbI 2 (s) 2. Describe the color of each aqueous solution of reactants lead nitrate and potassium iodide: Lead nitrate is yellow Potassium Iodine is white 3. Describe the color and state of each product (note, there are two products you will need to describe the color and state of). Recall, by “state” we mean aqueous (aq), gas (g), liquid (l), and solid (s). The color of PbI 2 is yellow. Its state is solid. The solid state of KNO 3 is white. Its state is solid. 1. Balance the following chemical equation: _2__FeBr 3 + _3___H 2 SO 4 _____Fe 2 (SO 4 ) 3 + __6__HBr 2. Balance the following chemical equation: ___C 7 H 16 + __11__O 2 __7___CO 2 + ___8___H 2 O 3. What is the formula weight (in amu) for ammonia, NH 3 ? Page 6 of 9
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Name(s): __________________ CHM 1020 W5 Molecular weight of Nitrogen is 14.01 g/ mol Molecular weight of Hydrogen is 1.01 g/mol NH3 has 1 N and 3 H Therefore the total molecular weight is 14.01 + ( 3*1.01) = 17.04 g/ mol 17.04 amu 4. How many molecules are present in one mole of water, H 2 O? Express your answer in scientific notation. 6.022 x 10^23 5. A person drinks 1.50 × 10 3 g of water, H 2 O, per day. How many moles is this? It is given that Mass of water given = 1.50x10 3 g Also, we know that Molar mass of water = 18g/mol Number of moles = given mass/molar mass Number of moles = 1.50x10 3 /18 15000/18 = 83.3 moles 6. How many grams of CH 3 OH are in 0.42 mol CH 3 OH? Molar mass of CH3OH= 32 g/mol therefore, 1 mol CH3OH= 32 g of CH3OH => 0.42 mol CH3OH= 0.42 * 32 = 13.44 g of CH3OH The mass is 13.44g of CH3 OH 7. The balanced equation N 2 ( g ) + 3 H 2 ( g ) 2 N H 3 ( g ) means which of the following? a. One gram of nitrogen reacts with three grams of hydrogen to form two grams of ammonia b. One mole of nitrogen reacts with three moles of hydrogen to form two moles of ammonia c. The equation means both a and b d. The equation means neither a nor b 8. Most of the industrial ammonia today is produced via the Haber Process. The Chemical reaction is one we’ve seen many times now: Page 7 of 9
Name(s): __________________ CHM 1020 W5 N 2 ( g ) + 3 H 2 ( g ) 2 N H 3 ( g ) If 20 moles of ammonia ( N H 3 ) are needed for a particular process, how many moles of diatomic Nitrogen ( N 2 ) are needed? 20 moles NH3 × (1 mole N2 / 2 moles NH3) = 10 moles N2 Practice with Stoichiometry Calculations – Problem Solving Workflow : 8. How many grams of H 2 are required to produce 7.50 grams of ammonia NH 3 ? Use the balanced equation (this time the equation is already balanced): N 2 ( g ) + 3 H 2 ( g ) 2 N H 3 ( g ) N 2 + 3H 2 --------> 2NH 3 molar mass of NH 3 = 17.031 g mol -1 mass of 2 mol of NH 3 = 17.031 g mol -1 x 2 mol = 34.062 g molar mass of H 2 = 2.015 g mol -1 mass of 3 mol of H 2 = 2.015 g mol -1 x 3 mol = 6.045 g To produce 34.062 g of NH 3 , 6.045 g of H 2 is used. To produce 7.50 g of NH 3 , the H2 needed is: = (6.045 g/34.062 g) x 7.50 g = 0.17747 x 7.50 g =1.33 g of H 2 is needed . Page 8 of 9
Name(s): __________________ CHM 1020 W5 9. If 54.7 grams of propane (C 3 H 8 ) and 89.6 grams of oxygen (O 2 ) are available in the balanced combustion reaction below: C 3 H 8 + 5 O 2 3 CO 2 + 4 H 2 O a) Which reactant is limiting (show your work)? O2 C 3 H 8 + 5O 2 =====> 3CO 2 + 4H 2 O Moles of C 3 H 8 mass given = 54.7 g molar mass of propane = 44.1 g/mol moles of propane = mass (g)/molar mass = 54.7 g/44.1 g/mol = 1.24 mol moles of Oxygen (O 2 ) mass given = 89.6 g molar mass of O 2 = 32.0 g/mol moles of O 2 = mass (g)/molar mass = 89.6 g/32.0 g/mol = 2.80 mol Moles Of CO2 from Propane moles of CO2 formed = 1.24 mol C3H8 (3 mol CO2/1 mol C3H8) = 3.72 mol CO2 Moles of CO2 from O2 moles of O2 = 2.80 mol moles of CO2 = 2.80 mol O2 (3 mol CO2/5 mol O2) = 1.68 mol CO2 b) What is the theoretical yield of C O 2 in grams? moles of CO2 = 1.68 mol molar mass of CO2 = 44.01 g/mol mass of CO2 produced = moles x molar mass = 1.68 mol x 44.01 g/mol = 73.9 g theoretical yield of CO2 is 73.9 g. Page 9 of 9
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